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Find the total amount of loans offered by each bank branch. | CREATE TABLE loan (branch_id VARCHAR); CREATE TABLE bank (bname VARCHAR, branch_id VARCHAR) | SELECT SUM(amount), T1.bname FROM bank AS T1 JOIN loan AS T2 ON T1.branch_id = T2.branch_id GROUP BY T1.bname | ### Context: CREATE TABLE loan (branch_id VARCHAR); CREATE TABLE bank (bname VARCHAR, branch_id VARCHAR) ### Question: Find the total amount of loans offered by each bank branch. ### Answer: SELECT SUM(amount), T1.bname FROM bank AS T1 JOIN loan AS T2 ON T1.branch_id = T2.branch_id GROUP BY T1.bname |
Find the name of customers who have more than one loan. | CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR); CREATE TABLE loan (cust_id VARCHAR) | SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name HAVING COUNT(*) > 1 | ### Context: CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR); CREATE TABLE loan (cust_id VARCHAR) ### Question: Find the name of customers who have more than one loan. ### Answer: SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name HAVING COUNT(*) > 1 |
Find the name and account balance of the customers who have loans with a total amount of more than 5000. | CREATE TABLE loan (cust_id VARCHAR, amount INTEGER); CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR, cust_id VARCHAR) | SELECT T1.cust_name, T1.acc_type FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name HAVING SUM(T2.amount) > 5000 | ### Context: CREATE TABLE loan (cust_id VARCHAR, amount INTEGER); CREATE TABLE customer (cust_name VARCHAR, acc_type VARCHAR, cust_id VARCHAR) ### Question: Find the name and account balance of the customers who have loans with a total amount of more than 5000. ### Answer: SELECT T1.cust_name, T1.acc_type FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id GROUP BY T1.cust_name HAVING SUM(T2.amount) > 5000 |
Find the name of bank branch that provided the greatest total amount of loans. | CREATE TABLE loan (branch_id VARCHAR, amount INTEGER); CREATE TABLE bank (bname VARCHAR, branch_id VARCHAR) | SELECT T1.bname FROM bank AS T1 JOIN loan AS T2 ON T1.branch_id = T2.branch_id GROUP BY T1.bname ORDER BY SUM(T2.amount) DESC LIMIT 1 | ### Context: CREATE TABLE loan (branch_id VARCHAR, amount INTEGER); CREATE TABLE bank (bname VARCHAR, branch_id VARCHAR) ### Question: Find the name of bank branch that provided the greatest total amount of loans. ### Answer: SELECT T1.bname FROM bank AS T1 JOIN loan AS T2 ON T1.branch_id = T2.branch_id GROUP BY T1.bname ORDER BY SUM(T2.amount) DESC LIMIT 1 |
Find the name of bank branch that provided the greatest total amount of loans to customers with credit score is less than 100. | CREATE TABLE loan (branch_id VARCHAR, cust_id VARCHAR, amount INTEGER); CREATE TABLE bank (bname VARCHAR, branch_id VARCHAR); CREATE TABLE customer (cust_id VARCHAR, credit_score INTEGER) | SELECT T2.bname FROM loan AS T1 JOIN bank AS T2 ON T1.branch_id = T2.branch_id JOIN customer AS T3 ON T1.cust_id = T3.cust_id WHERE T3.credit_score < 100 GROUP BY T2.bname ORDER BY SUM(T1.amount) DESC LIMIT 1 | ### Context: CREATE TABLE loan (branch_id VARCHAR, cust_id VARCHAR, amount INTEGER); CREATE TABLE bank (bname VARCHAR, branch_id VARCHAR); CREATE TABLE customer (cust_id VARCHAR, credit_score INTEGER) ### Question: Find the name of bank branch that provided the greatest total amount of loans to customers with credit score is less than 100. ### Answer: SELECT T2.bname FROM loan AS T1 JOIN bank AS T2 ON T1.branch_id = T2.branch_id JOIN customer AS T3 ON T1.cust_id = T3.cust_id WHERE T3.credit_score < 100 GROUP BY T2.bname ORDER BY SUM(T1.amount) DESC LIMIT 1 |
Find the name of bank branches that provided some loans. | CREATE TABLE loan (branch_id VARCHAR); CREATE TABLE bank (bname VARCHAR, branch_id VARCHAR) | SELECT DISTINCT T1.bname FROM bank AS T1 JOIN loan AS T2 ON T1.branch_id = T2.branch_id | ### Context: CREATE TABLE loan (branch_id VARCHAR); CREATE TABLE bank (bname VARCHAR, branch_id VARCHAR) ### Question: Find the name of bank branches that provided some loans. ### Answer: SELECT DISTINCT T1.bname FROM bank AS T1 JOIN loan AS T2 ON T1.branch_id = T2.branch_id |
Find the name and credit score of the customers who have some loans. | CREATE TABLE customer (cust_name VARCHAR, credit_score VARCHAR, cust_id VARCHAR); CREATE TABLE loan (cust_id VARCHAR) | SELECT DISTINCT T1.cust_name, T1.credit_score FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id | ### Context: CREATE TABLE customer (cust_name VARCHAR, credit_score VARCHAR, cust_id VARCHAR); CREATE TABLE loan (cust_id VARCHAR) ### Question: Find the name and credit score of the customers who have some loans. ### Answer: SELECT DISTINCT T1.cust_name, T1.credit_score FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id |
Find the the name of the customers who have a loan with amount more than 3000. | CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR); CREATE TABLE loan (cust_id VARCHAR) | SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE amount > 3000 | ### Context: CREATE TABLE customer (cust_name VARCHAR, cust_id VARCHAR); CREATE TABLE loan (cust_id VARCHAR) ### Question: Find the the name of the customers who have a loan with amount more than 3000. ### Answer: SELECT T1.cust_name FROM customer AS T1 JOIN loan AS T2 ON T1.cust_id = T2.cust_id WHERE amount > 3000 |
Find the city and name of bank branches that provide business loans. | CREATE TABLE bank (bname VARCHAR, city VARCHAR, branch_id VARCHAR); CREATE TABLE loan (branch_id VARCHAR, loan_type VARCHAR) | SELECT T1.bname, T1.city FROM bank AS T1 JOIN loan AS T2 ON T1.branch_id = T2.branch_id WHERE T2.loan_type = 'Business' | ### Context: CREATE TABLE bank (bname VARCHAR, city VARCHAR, branch_id VARCHAR); CREATE TABLE loan (branch_id VARCHAR, loan_type VARCHAR) ### Question: Find the city and name of bank branches that provide business loans. ### Answer: SELECT T1.bname, T1.city FROM bank AS T1 JOIN loan AS T2 ON T1.branch_id = T2.branch_id WHERE T2.loan_type = 'Business' |
Find the names of bank branches that have provided a loan to any customer whose credit score is below 100. | CREATE TABLE bank (bname VARCHAR, branch_id VARCHAR); CREATE TABLE customer (cust_id VARCHAR, credit_score INTEGER); CREATE TABLE loan (branch_id VARCHAR, cust_id VARCHAR) | SELECT T2.bname FROM loan AS T1 JOIN bank AS T2 ON T1.branch_id = T2.branch_id JOIN customer AS T3 ON T1.cust_id = T3.cust_id WHERE T3.credit_score < 100 | ### Context: CREATE TABLE bank (bname VARCHAR, branch_id VARCHAR); CREATE TABLE customer (cust_id VARCHAR, credit_score INTEGER); CREATE TABLE loan (branch_id VARCHAR, cust_id VARCHAR) ### Question: Find the names of bank branches that have provided a loan to any customer whose credit score is below 100. ### Answer: SELECT T2.bname FROM loan AS T1 JOIN bank AS T2 ON T1.branch_id = T2.branch_id JOIN customer AS T3 ON T1.cust_id = T3.cust_id WHERE T3.credit_score < 100 |
Find the total amount of loans provided by bank branches in the state of New York. | CREATE TABLE bank (branch_id VARCHAR, state VARCHAR); CREATE TABLE loan (amount INTEGER, branch_id VARCHAR) | SELECT SUM(T2.amount) FROM bank AS T1 JOIN loan AS T2 ON T1.branch_id = T2.branch_id WHERE T1.state = 'New York' | ### Context: CREATE TABLE bank (branch_id VARCHAR, state VARCHAR); CREATE TABLE loan (amount INTEGER, branch_id VARCHAR) ### Question: Find the total amount of loans provided by bank branches in the state of New York. ### Answer: SELECT SUM(T2.amount) FROM bank AS T1 JOIN loan AS T2 ON T1.branch_id = T2.branch_id WHERE T1.state = 'New York' |
Find the average credit score of the customers who have some loan. | CREATE TABLE loan (credit_score INTEGER, cust_id VARCHAR); CREATE TABLE customer (credit_score INTEGER, cust_id VARCHAR) | SELECT AVG(credit_score) FROM customer WHERE cust_id IN (SELECT cust_id FROM loan) | ### Context: CREATE TABLE loan (credit_score INTEGER, cust_id VARCHAR); CREATE TABLE customer (credit_score INTEGER, cust_id VARCHAR) ### Question: Find the average credit score of the customers who have some loan. ### Answer: SELECT AVG(credit_score) FROM customer WHERE cust_id IN (SELECT cust_id FROM loan) |
Find the average credit score of the customers who do not have any loan. | CREATE TABLE loan (credit_score INTEGER, cust_id VARCHAR); CREATE TABLE customer (credit_score INTEGER, cust_id VARCHAR) | SELECT AVG(credit_score) FROM customer WHERE NOT cust_id IN (SELECT cust_id FROM loan) | ### Context: CREATE TABLE loan (credit_score INTEGER, cust_id VARCHAR); CREATE TABLE customer (credit_score INTEGER, cust_id VARCHAR) ### Question: Find the average credit score of the customers who do not have any loan. ### Answer: SELECT AVG(credit_score) FROM customer WHERE NOT cust_id IN (SELECT cust_id FROM loan) |
How many assessment notes are there in total? | CREATE TABLE ASSESSMENT_NOTES (Id VARCHAR) | SELECT COUNT(*) FROM ASSESSMENT_NOTES | ### Context: CREATE TABLE ASSESSMENT_NOTES (Id VARCHAR) ### Question: How many assessment notes are there in total? ### Answer: SELECT COUNT(*) FROM ASSESSMENT_NOTES |
What are the dates of the assessment notes? | CREATE TABLE Assessment_Notes (date_of_notes VARCHAR) | SELECT date_of_notes FROM Assessment_Notes | ### Context: CREATE TABLE Assessment_Notes (date_of_notes VARCHAR) ### Question: What are the dates of the assessment notes? ### Answer: SELECT date_of_notes FROM Assessment_Notes |
How many addresses have zip code 197? | CREATE TABLE ADDRESSES (zip_postcode VARCHAR) | SELECT COUNT(*) FROM ADDRESSES WHERE zip_postcode = "197" | ### Context: CREATE TABLE ADDRESSES (zip_postcode VARCHAR) ### Question: How many addresses have zip code 197? ### Answer: SELECT COUNT(*) FROM ADDRESSES WHERE zip_postcode = "197" |
How many distinct incident type codes are there? | CREATE TABLE Behavior_Incident (incident_type_code VARCHAR) | SELECT COUNT(DISTINCT incident_type_code) FROM Behavior_Incident | ### Context: CREATE TABLE Behavior_Incident (incident_type_code VARCHAR) ### Question: How many distinct incident type codes are there? ### Answer: SELECT COUNT(DISTINCT incident_type_code) FROM Behavior_Incident |
Return all distinct detention type codes. | CREATE TABLE Detention (detention_type_code VARCHAR) | SELECT DISTINCT detention_type_code FROM Detention | ### Context: CREATE TABLE Detention (detention_type_code VARCHAR) ### Question: Return all distinct detention type codes. ### Answer: SELECT DISTINCT detention_type_code FROM Detention |
What are the start and end dates for incidents with incident type code "NOISE"? | CREATE TABLE Behavior_Incident (date_incident_start VARCHAR, date_incident_end VARCHAR, incident_type_code VARCHAR) | SELECT date_incident_start, date_incident_end FROM Behavior_Incident WHERE incident_type_code = "NOISE" | ### Context: CREATE TABLE Behavior_Incident (date_incident_start VARCHAR, date_incident_end VARCHAR, incident_type_code VARCHAR) ### Question: What are the start and end dates for incidents with incident type code "NOISE"? ### Answer: SELECT date_incident_start, date_incident_end FROM Behavior_Incident WHERE incident_type_code = "NOISE" |
Return all detention summaries. | CREATE TABLE Detention (detention_summary VARCHAR) | SELECT detention_summary FROM Detention | ### Context: CREATE TABLE Detention (detention_summary VARCHAR) ### Question: Return all detention summaries. ### Answer: SELECT detention_summary FROM Detention |
Return the cell phone number and email address for all students. | CREATE TABLE STUDENTS (cell_mobile_number VARCHAR, email_address VARCHAR) | SELECT cell_mobile_number, email_address FROM STUDENTS | ### Context: CREATE TABLE STUDENTS (cell_mobile_number VARCHAR, email_address VARCHAR) ### Question: Return the cell phone number and email address for all students. ### Answer: SELECT cell_mobile_number, email_address FROM STUDENTS |
What is the email of the student with first name "Emma" and last name "Rohan"? | CREATE TABLE Students (email_address VARCHAR, first_name VARCHAR, last_name VARCHAR) | SELECT email_address FROM Students WHERE first_name = "Emma" AND last_name = "Rohan" | ### Context: CREATE TABLE Students (email_address VARCHAR, first_name VARCHAR, last_name VARCHAR) ### Question: What is the email of the student with first name "Emma" and last name "Rohan"? ### Answer: SELECT email_address FROM Students WHERE first_name = "Emma" AND last_name = "Rohan" |
How many distinct students have been in detention? | CREATE TABLE Students_in_Detention (student_id VARCHAR) | SELECT COUNT(DISTINCT student_id) FROM Students_in_Detention | ### Context: CREATE TABLE Students_in_Detention (student_id VARCHAR) ### Question: How many distinct students have been in detention? ### Answer: SELECT COUNT(DISTINCT student_id) FROM Students_in_Detention |
What is the gender of the teacher with last name "Medhurst"? | CREATE TABLE TEACHERS (gender VARCHAR, last_name VARCHAR) | SELECT gender FROM TEACHERS WHERE last_name = "Medhurst" | ### Context: CREATE TABLE TEACHERS (gender VARCHAR, last_name VARCHAR) ### Question: What is the gender of the teacher with last name "Medhurst"? ### Answer: SELECT gender FROM TEACHERS WHERE last_name = "Medhurst" |
What is the incident type description for the incident type with code "VIOLENCE"? | CREATE TABLE Ref_Incident_Type (incident_type_description VARCHAR, incident_type_code VARCHAR) | SELECT incident_type_description FROM Ref_Incident_Type WHERE incident_type_code = "VIOLENCE" | ### Context: CREATE TABLE Ref_Incident_Type (incident_type_description VARCHAR, incident_type_code VARCHAR) ### Question: What is the incident type description for the incident type with code "VIOLENCE"? ### Answer: SELECT incident_type_description FROM Ref_Incident_Type WHERE incident_type_code = "VIOLENCE" |
Find the maximum and minimum monthly rental for all student addresses. | CREATE TABLE Student_Addresses (monthly_rental INTEGER) | SELECT MAX(monthly_rental), MIN(monthly_rental) FROM Student_Addresses | ### Context: CREATE TABLE Student_Addresses (monthly_rental INTEGER) ### Question: Find the maximum and minimum monthly rental for all student addresses. ### Answer: SELECT MAX(monthly_rental), MIN(monthly_rental) FROM Student_Addresses |
Find the first names of teachers whose email address contains the word "man". | CREATE TABLE Teachers (first_name VARCHAR, email_address VARCHAR) | SELECT first_name FROM Teachers WHERE email_address LIKE '%man%' | ### Context: CREATE TABLE Teachers (first_name VARCHAR, email_address VARCHAR) ### Question: Find the first names of teachers whose email address contains the word "man". ### Answer: SELECT first_name FROM Teachers WHERE email_address LIKE '%man%' |
List all information about the assessment notes sorted by date in ascending order. | CREATE TABLE Assessment_Notes (date_of_notes VARCHAR) | SELECT * FROM Assessment_Notes ORDER BY date_of_notes | ### Context: CREATE TABLE Assessment_Notes (date_of_notes VARCHAR) ### Question: List all information about the assessment notes sorted by date in ascending order. ### Answer: SELECT * FROM Assessment_Notes ORDER BY date_of_notes |
List all cities of addresses in alphabetical order. | CREATE TABLE Addresses (city VARCHAR) | SELECT city FROM Addresses ORDER BY city | ### Context: CREATE TABLE Addresses (city VARCHAR) ### Question: List all cities of addresses in alphabetical order. ### Answer: SELECT city FROM Addresses ORDER BY city |
Find the first names and last names of teachers in alphabetical order of last name. | CREATE TABLE Teachers (first_name VARCHAR, last_name VARCHAR) | SELECT first_name, last_name FROM Teachers ORDER BY last_name | ### Context: CREATE TABLE Teachers (first_name VARCHAR, last_name VARCHAR) ### Question: Find the first names and last names of teachers in alphabetical order of last name. ### Answer: SELECT first_name, last_name FROM Teachers ORDER BY last_name |
Find all information about student addresses, and sort by monthly rental in descending order. | CREATE TABLE Student_Addresses (monthly_rental VARCHAR) | SELECT * FROM Student_Addresses ORDER BY monthly_rental DESC | ### Context: CREATE TABLE Student_Addresses (monthly_rental VARCHAR) ### Question: Find all information about student addresses, and sort by monthly rental in descending order. ### Answer: SELECT * FROM Student_Addresses ORDER BY monthly_rental DESC |
Find the id and first name of the student that has the most number of assessment notes? | CREATE TABLE Students (first_name VARCHAR, student_id VARCHAR); CREATE TABLE Assessment_Notes (student_id VARCHAR) | SELECT T1.student_id, T2.first_name FROM Assessment_Notes AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id GROUP BY T1.student_id ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE Students (first_name VARCHAR, student_id VARCHAR); CREATE TABLE Assessment_Notes (student_id VARCHAR) ### Question: Find the id and first name of the student that has the most number of assessment notes? ### Answer: SELECT T1.student_id, T2.first_name FROM Assessment_Notes AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id GROUP BY T1.student_id ORDER BY COUNT(*) DESC LIMIT 1 |
Find the ids and first names of the 3 teachers that have the most number of assessment notes? | CREATE TABLE Assessment_Notes (teacher_id VARCHAR); CREATE TABLE Teachers (first_name VARCHAR, teacher_id VARCHAR) | SELECT T1.teacher_id, T2.first_name FROM Assessment_Notes AS T1 JOIN Teachers AS T2 ON T1.teacher_id = T2.teacher_id GROUP BY T1.teacher_id ORDER BY COUNT(*) DESC LIMIT 3 | ### Context: CREATE TABLE Assessment_Notes (teacher_id VARCHAR); CREATE TABLE Teachers (first_name VARCHAR, teacher_id VARCHAR) ### Question: Find the ids and first names of the 3 teachers that have the most number of assessment notes? ### Answer: SELECT T1.teacher_id, T2.first_name FROM Assessment_Notes AS T1 JOIN Teachers AS T2 ON T1.teacher_id = T2.teacher_id GROUP BY T1.teacher_id ORDER BY COUNT(*) DESC LIMIT 3 |
Find the id and last name of the student that has the most behavior incidents? | CREATE TABLE Students (last_name VARCHAR, student_id VARCHAR); CREATE TABLE Behavior_Incident (student_id VARCHAR) | SELECT T1.student_id, T2.last_name FROM Behavior_Incident AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id GROUP BY T1.student_id ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE Students (last_name VARCHAR, student_id VARCHAR); CREATE TABLE Behavior_Incident (student_id VARCHAR) ### Question: Find the id and last name of the student that has the most behavior incidents? ### Answer: SELECT T1.student_id, T2.last_name FROM Behavior_Incident AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id GROUP BY T1.student_id ORDER BY COUNT(*) DESC LIMIT 1 |
Find the id and last name of the teacher that has the most detentions with detention type code "AFTER"? | CREATE TABLE Detention (teacher_id VARCHAR, detention_type_code VARCHAR); CREATE TABLE Teachers (last_name VARCHAR, teacher_id VARCHAR) | SELECT T1.teacher_id, T2.last_name FROM Detention AS T1 JOIN Teachers AS T2 ON T1.teacher_id = T2.teacher_id WHERE T1.detention_type_code = "AFTER" GROUP BY T1.teacher_id ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE Detention (teacher_id VARCHAR, detention_type_code VARCHAR); CREATE TABLE Teachers (last_name VARCHAR, teacher_id VARCHAR) ### Question: Find the id and last name of the teacher that has the most detentions with detention type code "AFTER"? ### Answer: SELECT T1.teacher_id, T2.last_name FROM Detention AS T1 JOIN Teachers AS T2 ON T1.teacher_id = T2.teacher_id WHERE T1.detention_type_code = "AFTER" GROUP BY T1.teacher_id ORDER BY COUNT(*) DESC LIMIT 1 |
What are the id and first name of the student whose addresses have the highest average monthly rental? | CREATE TABLE Students (first_name VARCHAR, student_id VARCHAR); CREATE TABLE Student_Addresses (student_id VARCHAR) | SELECT T1.student_id, T2.first_name FROM Student_Addresses AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id GROUP BY T1.student_id ORDER BY AVG(monthly_rental) DESC LIMIT 1 | ### Context: CREATE TABLE Students (first_name VARCHAR, student_id VARCHAR); CREATE TABLE Student_Addresses (student_id VARCHAR) ### Question: What are the id and first name of the student whose addresses have the highest average monthly rental? ### Answer: SELECT T1.student_id, T2.first_name FROM Student_Addresses AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id GROUP BY T1.student_id ORDER BY AVG(monthly_rental) DESC LIMIT 1 |
Find the id and city of the student address with the highest average monthly rental. | CREATE TABLE Student_Addresses (address_id VARCHAR); CREATE TABLE Addresses (city VARCHAR, address_id VARCHAR) | SELECT T2.address_id, T1.city FROM Addresses AS T1 JOIN Student_Addresses AS T2 ON T1.address_id = T2.address_id GROUP BY T2.address_id ORDER BY AVG(monthly_rental) DESC LIMIT 1 | ### Context: CREATE TABLE Student_Addresses (address_id VARCHAR); CREATE TABLE Addresses (city VARCHAR, address_id VARCHAR) ### Question: Find the id and city of the student address with the highest average monthly rental. ### Answer: SELECT T2.address_id, T1.city FROM Addresses AS T1 JOIN Student_Addresses AS T2 ON T1.address_id = T2.address_id GROUP BY T2.address_id ORDER BY AVG(monthly_rental) DESC LIMIT 1 |
What are the code and description of the most frequent behavior incident type? | CREATE TABLE Ref_Incident_Type (incident_type_description VARCHAR, incident_type_code VARCHAR); CREATE TABLE Behavior_Incident (incident_type_code VARCHAR) | SELECT T1.incident_type_code, T2.incident_type_description FROM Behavior_Incident AS T1 JOIN Ref_Incident_Type AS T2 ON T1.incident_type_code = T2.incident_type_code GROUP BY T1.incident_type_code ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE Ref_Incident_Type (incident_type_description VARCHAR, incident_type_code VARCHAR); CREATE TABLE Behavior_Incident (incident_type_code VARCHAR) ### Question: What are the code and description of the most frequent behavior incident type? ### Answer: SELECT T1.incident_type_code, T2.incident_type_description FROM Behavior_Incident AS T1 JOIN Ref_Incident_Type AS T2 ON T1.incident_type_code = T2.incident_type_code GROUP BY T1.incident_type_code ORDER BY COUNT(*) DESC LIMIT 1 |
What are the code and description of the least frequent detention type ? | CREATE TABLE Ref_Detention_Type (detention_type_description VARCHAR, detention_type_code VARCHAR); CREATE TABLE Detention (detention_type_code VARCHAR) | SELECT T1.detention_type_code, T2.detention_type_description FROM Detention AS T1 JOIN Ref_Detention_Type AS T2 ON T1.detention_type_code = T2.detention_type_code GROUP BY T1.detention_type_code ORDER BY COUNT(*) LIMIT 1 | ### Context: CREATE TABLE Ref_Detention_Type (detention_type_description VARCHAR, detention_type_code VARCHAR); CREATE TABLE Detention (detention_type_code VARCHAR) ### Question: What are the code and description of the least frequent detention type ? ### Answer: SELECT T1.detention_type_code, T2.detention_type_description FROM Detention AS T1 JOIN Ref_Detention_Type AS T2 ON T1.detention_type_code = T2.detention_type_code GROUP BY T1.detention_type_code ORDER BY COUNT(*) LIMIT 1 |
Find the dates of assessment notes for students with first name "Fanny". | CREATE TABLE Students (student_id VARCHAR, first_name VARCHAR); CREATE TABLE Assessment_Notes (date_of_notes VARCHAR, student_id VARCHAR) | SELECT T1.date_of_notes FROM Assessment_Notes AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id WHERE T2.first_name = "Fanny" | ### Context: CREATE TABLE Students (student_id VARCHAR, first_name VARCHAR); CREATE TABLE Assessment_Notes (date_of_notes VARCHAR, student_id VARCHAR) ### Question: Find the dates of assessment notes for students with first name "Fanny". ### Answer: SELECT T1.date_of_notes FROM Assessment_Notes AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id WHERE T2.first_name = "Fanny" |
Find the texts of assessment notes for teachers with last name "Schuster". | CREATE TABLE Assessment_Notes (text_of_notes VARCHAR, teacher_id VARCHAR); CREATE TABLE Teachers (teacher_id VARCHAR, last_name VARCHAR) | SELECT T1.text_of_notes FROM Assessment_Notes AS T1 JOIN Teachers AS T2 ON T1.teacher_id = T2.teacher_id WHERE T2.last_name = "Schuster" | ### Context: CREATE TABLE Assessment_Notes (text_of_notes VARCHAR, teacher_id VARCHAR); CREATE TABLE Teachers (teacher_id VARCHAR, last_name VARCHAR) ### Question: Find the texts of assessment notes for teachers with last name "Schuster". ### Answer: SELECT T1.text_of_notes FROM Assessment_Notes AS T1 JOIN Teachers AS T2 ON T1.teacher_id = T2.teacher_id WHERE T2.last_name = "Schuster" |
Find the start and end dates of behavior incidents of students with last name "Fahey". | CREATE TABLE Behavior_Incident (date_incident_start VARCHAR, student_id VARCHAR); CREATE TABLE Students (student_id VARCHAR, last_name VARCHAR) | SELECT T1.date_incident_start, date_incident_end FROM Behavior_Incident AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id WHERE T2.last_name = "Fahey" | ### Context: CREATE TABLE Behavior_Incident (date_incident_start VARCHAR, student_id VARCHAR); CREATE TABLE Students (student_id VARCHAR, last_name VARCHAR) ### Question: Find the start and end dates of behavior incidents of students with last name "Fahey". ### Answer: SELECT T1.date_incident_start, date_incident_end FROM Behavior_Incident AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id WHERE T2.last_name = "Fahey" |
Find the start and end dates of detentions of teachers with last name "Schultz". | CREATE TABLE Detention (datetime_detention_start VARCHAR, teacher_id VARCHAR); CREATE TABLE Teachers (teacher_id VARCHAR, last_name VARCHAR) | SELECT T1.datetime_detention_start, datetime_detention_end FROM Detention AS T1 JOIN Teachers AS T2 ON T1.teacher_id = T2.teacher_id WHERE T2.last_name = "Schultz" | ### Context: CREATE TABLE Detention (datetime_detention_start VARCHAR, teacher_id VARCHAR); CREATE TABLE Teachers (teacher_id VARCHAR, last_name VARCHAR) ### Question: Find the start and end dates of detentions of teachers with last name "Schultz". ### Answer: SELECT T1.datetime_detention_start, datetime_detention_end FROM Detention AS T1 JOIN Teachers AS T2 ON T1.teacher_id = T2.teacher_id WHERE T2.last_name = "Schultz" |
What are the id and zip code of the address with the highest monthly rental? | CREATE TABLE Student_Addresses (address_id VARCHAR); CREATE TABLE Addresses (zip_postcode VARCHAR, address_id VARCHAR) | SELECT T2.address_id, T1.zip_postcode FROM Addresses AS T1 JOIN Student_Addresses AS T2 ON T1.address_id = T2.address_id ORDER BY monthly_rental DESC LIMIT 1 | ### Context: CREATE TABLE Student_Addresses (address_id VARCHAR); CREATE TABLE Addresses (zip_postcode VARCHAR, address_id VARCHAR) ### Question: What are the id and zip code of the address with the highest monthly rental? ### Answer: SELECT T2.address_id, T1.zip_postcode FROM Addresses AS T1 JOIN Student_Addresses AS T2 ON T1.address_id = T2.address_id ORDER BY monthly_rental DESC LIMIT 1 |
What is the cell phone number of the student whose address has the lowest monthly rental? | CREATE TABLE Students (cell_mobile_number VARCHAR, student_id VARCHAR); CREATE TABLE Student_Addresses (student_id VARCHAR, monthly_rental VARCHAR) | SELECT T2.cell_mobile_number FROM Student_Addresses AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id ORDER BY T1.monthly_rental LIMIT 1 | ### Context: CREATE TABLE Students (cell_mobile_number VARCHAR, student_id VARCHAR); CREATE TABLE Student_Addresses (student_id VARCHAR, monthly_rental VARCHAR) ### Question: What is the cell phone number of the student whose address has the lowest monthly rental? ### Answer: SELECT T2.cell_mobile_number FROM Student_Addresses AS T1 JOIN Students AS T2 ON T1.student_id = T2.student_id ORDER BY T1.monthly_rental LIMIT 1 |
What are the monthly rentals of student addresses in Texas state? | CREATE TABLE Addresses (address_id VARCHAR, state_province_county VARCHAR); CREATE TABLE Student_Addresses (monthly_rental VARCHAR, address_id VARCHAR) | SELECT T2.monthly_rental FROM Addresses AS T1 JOIN Student_Addresses AS T2 ON T1.address_id = T2.address_id WHERE T1.state_province_county = "Texas" | ### Context: CREATE TABLE Addresses (address_id VARCHAR, state_province_county VARCHAR); CREATE TABLE Student_Addresses (monthly_rental VARCHAR, address_id VARCHAR) ### Question: What are the monthly rentals of student addresses in Texas state? ### Answer: SELECT T2.monthly_rental FROM Addresses AS T1 JOIN Student_Addresses AS T2 ON T1.address_id = T2.address_id WHERE T1.state_province_county = "Texas" |
What are the first names and last names of students with address in Wisconsin state? | CREATE TABLE Addresses (address_id VARCHAR, state_province_county VARCHAR); CREATE TABLE Students (first_name VARCHAR, last_name VARCHAR, address_id VARCHAR) | SELECT T2.first_name, T2.last_name FROM Addresses AS T1 JOIN Students AS T2 ON T1.address_id = T2.address_id WHERE T1.state_province_county = "Wisconsin" | ### Context: CREATE TABLE Addresses (address_id VARCHAR, state_province_county VARCHAR); CREATE TABLE Students (first_name VARCHAR, last_name VARCHAR, address_id VARCHAR) ### Question: What are the first names and last names of students with address in Wisconsin state? ### Answer: SELECT T2.first_name, T2.last_name FROM Addresses AS T1 JOIN Students AS T2 ON T1.address_id = T2.address_id WHERE T1.state_province_county = "Wisconsin" |
What are the line 1 and average monthly rentals of all student addresses? | CREATE TABLE Addresses (line_1 VARCHAR, address_id VARCHAR); CREATE TABLE Student_Addresses (monthly_rental INTEGER, address_id VARCHAR) | SELECT T1.line_1, AVG(T2.monthly_rental) FROM Addresses AS T1 JOIN Student_Addresses AS T2 ON T1.address_id = T2.address_id GROUP BY T2.address_id | ### Context: CREATE TABLE Addresses (line_1 VARCHAR, address_id VARCHAR); CREATE TABLE Student_Addresses (monthly_rental INTEGER, address_id VARCHAR) ### Question: What are the line 1 and average monthly rentals of all student addresses? ### Answer: SELECT T1.line_1, AVG(T2.monthly_rental) FROM Addresses AS T1 JOIN Student_Addresses AS T2 ON T1.address_id = T2.address_id GROUP BY T2.address_id |
What is the zip code of the address where the teacher with first name "Lyla" lives? | CREATE TABLE Teachers (address_id VARCHAR, first_name VARCHAR); CREATE TABLE Addresses (zip_postcode VARCHAR, address_id VARCHAR) | SELECT T1.zip_postcode FROM Addresses AS T1 JOIN Teachers AS T2 ON T1.address_id = T2.address_id WHERE T2.first_name = "Lyla" | ### Context: CREATE TABLE Teachers (address_id VARCHAR, first_name VARCHAR); CREATE TABLE Addresses (zip_postcode VARCHAR, address_id VARCHAR) ### Question: What is the zip code of the address where the teacher with first name "Lyla" lives? ### Answer: SELECT T1.zip_postcode FROM Addresses AS T1 JOIN Teachers AS T2 ON T1.address_id = T2.address_id WHERE T2.first_name = "Lyla" |
What are the email addresses of teachers whose address has zip code "918"? | CREATE TABLE Addresses (address_id VARCHAR, zip_postcode VARCHAR); CREATE TABLE Teachers (email_address VARCHAR, address_id VARCHAR) | SELECT T2.email_address FROM Addresses AS T1 JOIN Teachers AS T2 ON T1.address_id = T2.address_id WHERE T1.zip_postcode = "918" | ### Context: CREATE TABLE Addresses (address_id VARCHAR, zip_postcode VARCHAR); CREATE TABLE Teachers (email_address VARCHAR, address_id VARCHAR) ### Question: What are the email addresses of teachers whose address has zip code "918"? ### Answer: SELECT T2.email_address FROM Addresses AS T1 JOIN Teachers AS T2 ON T1.address_id = T2.address_id WHERE T1.zip_postcode = "918" |
How many students are not involved in any behavior incident? | CREATE TABLE STUDENTS (student_id VARCHAR); CREATE TABLE Behavior_Incident (student_id VARCHAR) | SELECT COUNT(*) FROM STUDENTS WHERE NOT student_id IN (SELECT student_id FROM Behavior_Incident) | ### Context: CREATE TABLE STUDENTS (student_id VARCHAR); CREATE TABLE Behavior_Incident (student_id VARCHAR) ### Question: How many students are not involved in any behavior incident? ### Answer: SELECT COUNT(*) FROM STUDENTS WHERE NOT student_id IN (SELECT student_id FROM Behavior_Incident) |
Find the last names of teachers who are not involved in any detention. | CREATE TABLE Teachers (last_name VARCHAR); CREATE TABLE Teachers (last_name VARCHAR, teacher_id VARCHAR); CREATE TABLE Detention (teacher_id VARCHAR) | SELECT last_name FROM Teachers EXCEPT SELECT T1.last_name FROM Teachers AS T1 JOIN Detention AS T2 ON T1.teacher_id = T2.teacher_id | ### Context: CREATE TABLE Teachers (last_name VARCHAR); CREATE TABLE Teachers (last_name VARCHAR, teacher_id VARCHAR); CREATE TABLE Detention (teacher_id VARCHAR) ### Question: Find the last names of teachers who are not involved in any detention. ### Answer: SELECT last_name FROM Teachers EXCEPT SELECT T1.last_name FROM Teachers AS T1 JOIN Detention AS T2 ON T1.teacher_id = T2.teacher_id |
What are the line 1 of addresses shared by some students and some teachers? | CREATE TABLE Teachers (address_id VARCHAR); CREATE TABLE Students (address_id VARCHAR); CREATE TABLE Addresses (line_1 VARCHAR, address_id VARCHAR) | SELECT T1.line_1 FROM Addresses AS T1 JOIN Students AS T2 ON T1.address_id = T2.address_id INTERSECT SELECT T1.line_1 FROM Addresses AS T1 JOIN Teachers AS T2 ON T1.address_id = T2.address_id | ### Context: CREATE TABLE Teachers (address_id VARCHAR); CREATE TABLE Students (address_id VARCHAR); CREATE TABLE Addresses (line_1 VARCHAR, address_id VARCHAR) ### Question: What are the line 1 of addresses shared by some students and some teachers? ### Answer: SELECT T1.line_1 FROM Addresses AS T1 JOIN Students AS T2 ON T1.address_id = T2.address_id INTERSECT SELECT T1.line_1 FROM Addresses AS T1 JOIN Teachers AS T2 ON T1.address_id = T2.address_id |
Which assets have 2 parts and have less than 2 fault logs? List the asset id and detail. | CREATE TABLE Assets (asset_id VARCHAR, asset_details VARCHAR); CREATE TABLE Asset_Parts (asset_id VARCHAR); CREATE TABLE Fault_Log (asset_id VARCHAR) | SELECT T1.asset_id, T1.asset_details FROM Assets AS T1 JOIN Asset_Parts AS T2 ON T1.asset_id = T2.asset_id GROUP BY T1.asset_id HAVING COUNT(*) = 2 INTERSECT SELECT T1.asset_id, T1.asset_details FROM Assets AS T1 JOIN Fault_Log AS T2 ON T1.asset_id = T2.asset_id GROUP BY T1.asset_id HAVING COUNT(*) < 2 | ### Context: CREATE TABLE Assets (asset_id VARCHAR, asset_details VARCHAR); CREATE TABLE Asset_Parts (asset_id VARCHAR); CREATE TABLE Fault_Log (asset_id VARCHAR) ### Question: Which assets have 2 parts and have less than 2 fault logs? List the asset id and detail. ### Answer: SELECT T1.asset_id, T1.asset_details FROM Assets AS T1 JOIN Asset_Parts AS T2 ON T1.asset_id = T2.asset_id GROUP BY T1.asset_id HAVING COUNT(*) = 2 INTERSECT SELECT T1.asset_id, T1.asset_details FROM Assets AS T1 JOIN Fault_Log AS T2 ON T1.asset_id = T2.asset_id GROUP BY T1.asset_id HAVING COUNT(*) < 2 |
How many assets does each maintenance contract contain? List the number and the contract id. | CREATE TABLE Assets (maintenance_contract_id VARCHAR); CREATE TABLE Maintenance_Contracts (maintenance_contract_id VARCHAR) | SELECT COUNT(*), T1.maintenance_contract_id FROM Maintenance_Contracts AS T1 JOIN Assets AS T2 ON T1.maintenance_contract_id = T2.maintenance_contract_id GROUP BY T1.maintenance_contract_id | ### Context: CREATE TABLE Assets (maintenance_contract_id VARCHAR); CREATE TABLE Maintenance_Contracts (maintenance_contract_id VARCHAR) ### Question: How many assets does each maintenance contract contain? List the number and the contract id. ### Answer: SELECT COUNT(*), T1.maintenance_contract_id FROM Maintenance_Contracts AS T1 JOIN Assets AS T2 ON T1.maintenance_contract_id = T2.maintenance_contract_id GROUP BY T1.maintenance_contract_id |
How many assets does each third party company supply? List the count and the company id. | CREATE TABLE Assets (supplier_company_id VARCHAR); CREATE TABLE Third_Party_Companies (company_id VARCHAR) | SELECT COUNT(*), T1.company_id FROM Third_Party_Companies AS T1 JOIN Assets AS T2 ON T1.company_id = T2.supplier_company_id GROUP BY T1.company_id | ### Context: CREATE TABLE Assets (supplier_company_id VARCHAR); CREATE TABLE Third_Party_Companies (company_id VARCHAR) ### Question: How many assets does each third party company supply? List the count and the company id. ### Answer: SELECT COUNT(*), T1.company_id FROM Third_Party_Companies AS T1 JOIN Assets AS T2 ON T1.company_id = T2.supplier_company_id GROUP BY T1.company_id |
Which third party companies have at least 2 maintenance engineers or have at least 2 maintenance contracts? List the company id and name. | CREATE TABLE Maintenance_Contracts (maintenance_contract_company_id VARCHAR); CREATE TABLE Maintenance_Engineers (company_id VARCHAR); CREATE TABLE Third_Party_Companies (company_id VARCHAR, company_name VARCHAR) | SELECT T1.company_id, T1.company_name FROM Third_Party_Companies AS T1 JOIN Maintenance_Engineers AS T2 ON T1.company_id = T2.company_id GROUP BY T1.company_id HAVING COUNT(*) >= 2 UNION SELECT T3.company_id, T3.company_name FROM Third_Party_Companies AS T3 JOIN Maintenance_Contracts AS T4 ON T3.company_id = T4.maintenance_contract_company_id GROUP BY T3.company_id HAVING COUNT(*) >= 2 | ### Context: CREATE TABLE Maintenance_Contracts (maintenance_contract_company_id VARCHAR); CREATE TABLE Maintenance_Engineers (company_id VARCHAR); CREATE TABLE Third_Party_Companies (company_id VARCHAR, company_name VARCHAR) ### Question: Which third party companies have at least 2 maintenance engineers or have at least 2 maintenance contracts? List the company id and name. ### Answer: SELECT T1.company_id, T1.company_name FROM Third_Party_Companies AS T1 JOIN Maintenance_Engineers AS T2 ON T1.company_id = T2.company_id GROUP BY T1.company_id HAVING COUNT(*) >= 2 UNION SELECT T3.company_id, T3.company_name FROM Third_Party_Companies AS T3 JOIN Maintenance_Contracts AS T4 ON T3.company_id = T4.maintenance_contract_company_id GROUP BY T3.company_id HAVING COUNT(*) >= 2 |
What is the name and id of the staff who recorded the fault log but has not contacted any visiting engineers? | CREATE TABLE Engineer_Visits (contact_staff_id VARCHAR); CREATE TABLE Staff (staff_name VARCHAR, staff_id VARCHAR); CREATE TABLE Fault_Log (recorded_by_staff_id VARCHAR) | SELECT T1.staff_name, T1.staff_id FROM Staff AS T1 JOIN Fault_Log AS T2 ON T1.staff_id = T2.recorded_by_staff_id EXCEPT SELECT T3.staff_name, T3.staff_id FROM Staff AS T3 JOIN Engineer_Visits AS T4 ON T3.staff_id = T4.contact_staff_id | ### Context: CREATE TABLE Engineer_Visits (contact_staff_id VARCHAR); CREATE TABLE Staff (staff_name VARCHAR, staff_id VARCHAR); CREATE TABLE Fault_Log (recorded_by_staff_id VARCHAR) ### Question: What is the name and id of the staff who recorded the fault log but has not contacted any visiting engineers? ### Answer: SELECT T1.staff_name, T1.staff_id FROM Staff AS T1 JOIN Fault_Log AS T2 ON T1.staff_id = T2.recorded_by_staff_id EXCEPT SELECT T3.staff_name, T3.staff_id FROM Staff AS T3 JOIN Engineer_Visits AS T4 ON T3.staff_id = T4.contact_staff_id |
Which engineer has visited the most times? Show the engineer id, first name and last name. | CREATE TABLE Maintenance_Engineers (engineer_id VARCHAR, first_name VARCHAR, last_name VARCHAR); CREATE TABLE Engineer_Visits (Id VARCHAR) | SELECT T1.engineer_id, T1.first_name, T1.last_name FROM Maintenance_Engineers AS T1 JOIN Engineer_Visits AS T2 GROUP BY T1.engineer_id ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE Maintenance_Engineers (engineer_id VARCHAR, first_name VARCHAR, last_name VARCHAR); CREATE TABLE Engineer_Visits (Id VARCHAR) ### Question: Which engineer has visited the most times? Show the engineer id, first name and last name. ### Answer: SELECT T1.engineer_id, T1.first_name, T1.last_name FROM Maintenance_Engineers AS T1 JOIN Engineer_Visits AS T2 GROUP BY T1.engineer_id ORDER BY COUNT(*) DESC LIMIT 1 |
Which parts have more than 2 faults? Show the part name and id. | CREATE TABLE Parts (part_name VARCHAR, part_id VARCHAR); CREATE TABLE Part_Faults (part_id VARCHAR) | SELECT T1.part_name, T1.part_id FROM Parts AS T1 JOIN Part_Faults AS T2 ON T1.part_id = T2.part_id GROUP BY T1.part_id HAVING COUNT(*) > 2 | ### Context: CREATE TABLE Parts (part_name VARCHAR, part_id VARCHAR); CREATE TABLE Part_Faults (part_id VARCHAR) ### Question: Which parts have more than 2 faults? Show the part name and id. ### Answer: SELECT T1.part_name, T1.part_id FROM Parts AS T1 JOIN Part_Faults AS T2 ON T1.part_id = T2.part_id GROUP BY T1.part_id HAVING COUNT(*) > 2 |
List all every engineer's first name, last name, details and coresponding skill description. | CREATE TABLE Maintenance_Engineers (first_name VARCHAR, last_name VARCHAR, other_details VARCHAR, engineer_id VARCHAR); CREATE TABLE Engineer_Skills (engineer_id VARCHAR, skill_id VARCHAR); CREATE TABLE Skills (skill_description VARCHAR, skill_id VARCHAR) | SELECT T1.first_name, T1.last_name, T1.other_details, T3.skill_description FROM Maintenance_Engineers AS T1 JOIN Engineer_Skills AS T2 ON T1.engineer_id = T2.engineer_id JOIN Skills AS T3 ON T2.skill_id = T3.skill_id | ### Context: CREATE TABLE Maintenance_Engineers (first_name VARCHAR, last_name VARCHAR, other_details VARCHAR, engineer_id VARCHAR); CREATE TABLE Engineer_Skills (engineer_id VARCHAR, skill_id VARCHAR); CREATE TABLE Skills (skill_description VARCHAR, skill_id VARCHAR) ### Question: List all every engineer's first name, last name, details and coresponding skill description. ### Answer: SELECT T1.first_name, T1.last_name, T1.other_details, T3.skill_description FROM Maintenance_Engineers AS T1 JOIN Engineer_Skills AS T2 ON T1.engineer_id = T2.engineer_id JOIN Skills AS T3 ON T2.skill_id = T3.skill_id |
For all the faults of different parts, what are all the decriptions of the skills required to fix them? List the name of the faults and the skill description. | CREATE TABLE Skills_Required_To_Fix (part_fault_id VARCHAR, skill_id VARCHAR); CREATE TABLE Part_Faults (fault_short_name VARCHAR, part_fault_id VARCHAR); CREATE TABLE Skills (skill_description VARCHAR, skill_id VARCHAR) | SELECT T1.fault_short_name, T3.skill_description FROM Part_Faults AS T1 JOIN Skills_Required_To_Fix AS T2 ON T1.part_fault_id = T2.part_fault_id JOIN Skills AS T3 ON T2.skill_id = T3.skill_id | ### Context: CREATE TABLE Skills_Required_To_Fix (part_fault_id VARCHAR, skill_id VARCHAR); CREATE TABLE Part_Faults (fault_short_name VARCHAR, part_fault_id VARCHAR); CREATE TABLE Skills (skill_description VARCHAR, skill_id VARCHAR) ### Question: For all the faults of different parts, what are all the decriptions of the skills required to fix them? List the name of the faults and the skill description. ### Answer: SELECT T1.fault_short_name, T3.skill_description FROM Part_Faults AS T1 JOIN Skills_Required_To_Fix AS T2 ON T1.part_fault_id = T2.part_fault_id JOIN Skills AS T3 ON T2.skill_id = T3.skill_id |
How many assets can each parts be used in? List the part name and the number. | CREATE TABLE Parts (part_name VARCHAR, part_id VARCHAR); CREATE TABLE Asset_Parts (part_id VARCHAR) | SELECT T1.part_name, COUNT(*) FROM Parts AS T1 JOIN Asset_Parts AS T2 ON T1.part_id = T2.part_id GROUP BY T1.part_name | ### Context: CREATE TABLE Parts (part_name VARCHAR, part_id VARCHAR); CREATE TABLE Asset_Parts (part_id VARCHAR) ### Question: How many assets can each parts be used in? List the part name and the number. ### Answer: SELECT T1.part_name, COUNT(*) FROM Parts AS T1 JOIN Asset_Parts AS T2 ON T1.part_id = T2.part_id GROUP BY T1.part_name |
What are all the fault descriptions and the fault status of all the faults recoreded in the logs? | CREATE TABLE Fault_Log (fault_description VARCHAR, fault_log_entry_id VARCHAR); CREATE TABLE Fault_Log_Parts (fault_status VARCHAR, fault_log_entry_id VARCHAR) | SELECT T1.fault_description, T2.fault_status FROM Fault_Log AS T1 JOIN Fault_Log_Parts AS T2 ON T1.fault_log_entry_id = T2.fault_log_entry_id | ### Context: CREATE TABLE Fault_Log (fault_description VARCHAR, fault_log_entry_id VARCHAR); CREATE TABLE Fault_Log_Parts (fault_status VARCHAR, fault_log_entry_id VARCHAR) ### Question: What are all the fault descriptions and the fault status of all the faults recoreded in the logs? ### Answer: SELECT T1.fault_description, T2.fault_status FROM Fault_Log AS T1 JOIN Fault_Log_Parts AS T2 ON T1.fault_log_entry_id = T2.fault_log_entry_id |
How many engineer visits are required at most for a single fault log? List the number and the log entry id. | CREATE TABLE Fault_Log (fault_log_entry_id VARCHAR); CREATE TABLE Engineer_Visits (fault_log_entry_id VARCHAR) | SELECT COUNT(*), T1.fault_log_entry_id FROM Fault_Log AS T1 JOIN Engineer_Visits AS T2 ON T1.fault_log_entry_id = T2.fault_log_entry_id GROUP BY T1.fault_log_entry_id ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE Fault_Log (fault_log_entry_id VARCHAR); CREATE TABLE Engineer_Visits (fault_log_entry_id VARCHAR) ### Question: How many engineer visits are required at most for a single fault log? List the number and the log entry id. ### Answer: SELECT COUNT(*), T1.fault_log_entry_id FROM Fault_Log AS T1 JOIN Engineer_Visits AS T2 ON T1.fault_log_entry_id = T2.fault_log_entry_id GROUP BY T1.fault_log_entry_id ORDER BY COUNT(*) DESC LIMIT 1 |
What are all the distinct last names of all the engineers? | CREATE TABLE Maintenance_Engineers (last_name VARCHAR) | SELECT DISTINCT last_name FROM Maintenance_Engineers | ### Context: CREATE TABLE Maintenance_Engineers (last_name VARCHAR) ### Question: What are all the distinct last names of all the engineers? ### Answer: SELECT DISTINCT last_name FROM Maintenance_Engineers |
How many fault status codes are recorded in the fault log parts table? | CREATE TABLE Fault_Log_Parts (fault_status VARCHAR) | SELECT DISTINCT fault_status FROM Fault_Log_Parts | ### Context: CREATE TABLE Fault_Log_Parts (fault_status VARCHAR) ### Question: How many fault status codes are recorded in the fault log parts table? ### Answer: SELECT DISTINCT fault_status FROM Fault_Log_Parts |
Which engineers have never visited to maintain the assets? List the engineer first name and last name. | CREATE TABLE Engineer_Visits (first_name VARCHAR, last_name VARCHAR, engineer_id VARCHAR); CREATE TABLE Maintenance_Engineers (first_name VARCHAR, last_name VARCHAR, engineer_id VARCHAR) | SELECT first_name, last_name FROM Maintenance_Engineers WHERE NOT engineer_id IN (SELECT engineer_id FROM Engineer_Visits) | ### Context: CREATE TABLE Engineer_Visits (first_name VARCHAR, last_name VARCHAR, engineer_id VARCHAR); CREATE TABLE Maintenance_Engineers (first_name VARCHAR, last_name VARCHAR, engineer_id VARCHAR) ### Question: Which engineers have never visited to maintain the assets? List the engineer first name and last name. ### Answer: SELECT first_name, last_name FROM Maintenance_Engineers WHERE NOT engineer_id IN (SELECT engineer_id FROM Engineer_Visits) |
List the asset id, details, make and model for every asset. | CREATE TABLE Assets (asset_id VARCHAR, asset_details VARCHAR, asset_make VARCHAR, asset_model VARCHAR) | SELECT asset_id, asset_details, asset_make, asset_model FROM Assets | ### Context: CREATE TABLE Assets (asset_id VARCHAR, asset_details VARCHAR, asset_make VARCHAR, asset_model VARCHAR) ### Question: List the asset id, details, make and model for every asset. ### Answer: SELECT asset_id, asset_details, asset_make, asset_model FROM Assets |
When was the first asset acquired? | CREATE TABLE Assets (asset_acquired_date VARCHAR) | SELECT asset_acquired_date FROM Assets ORDER BY asset_acquired_date LIMIT 1 | ### Context: CREATE TABLE Assets (asset_acquired_date VARCHAR) ### Question: When was the first asset acquired? ### Answer: SELECT asset_acquired_date FROM Assets ORDER BY asset_acquired_date LIMIT 1 |
Which part fault requires the most number of skills to fix? List part id and name. | CREATE TABLE Part_Faults (part_id VARCHAR, part_fault_id VARCHAR); CREATE TABLE Skills_Required_To_Fix (part_fault_id VARCHAR); CREATE TABLE Parts (part_id VARCHAR, part_name VARCHAR) | SELECT T1.part_id, T1.part_name FROM Parts AS T1 JOIN Part_Faults AS T2 ON T1.part_id = T2.part_id JOIN Skills_Required_To_Fix AS T3 ON T2.part_fault_id = T3.part_fault_id GROUP BY T1.part_id ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE Part_Faults (part_id VARCHAR, part_fault_id VARCHAR); CREATE TABLE Skills_Required_To_Fix (part_fault_id VARCHAR); CREATE TABLE Parts (part_id VARCHAR, part_name VARCHAR) ### Question: Which part fault requires the most number of skills to fix? List part id and name. ### Answer: SELECT T1.part_id, T1.part_name FROM Parts AS T1 JOIN Part_Faults AS T2 ON T1.part_id = T2.part_id JOIN Skills_Required_To_Fix AS T3 ON T2.part_fault_id = T3.part_fault_id GROUP BY T1.part_id ORDER BY COUNT(*) DESC LIMIT 1 |
Which kind of part has the least number of faults? List the part name. | CREATE TABLE Parts (part_name VARCHAR, part_id VARCHAR); CREATE TABLE Part_Faults (part_id VARCHAR) | SELECT T1.part_name FROM Parts AS T1 JOIN Part_Faults AS T2 ON T1.part_id = T2.part_id GROUP BY T1.part_name ORDER BY COUNT(*) LIMIT 1 | ### Context: CREATE TABLE Parts (part_name VARCHAR, part_id VARCHAR); CREATE TABLE Part_Faults (part_id VARCHAR) ### Question: Which kind of part has the least number of faults? List the part name. ### Answer: SELECT T1.part_name FROM Parts AS T1 JOIN Part_Faults AS T2 ON T1.part_id = T2.part_id GROUP BY T1.part_name ORDER BY COUNT(*) LIMIT 1 |
Among those engineers who have visited, which engineer makes the least number of visits? List the engineer id, first name and last name. | CREATE TABLE Engineer_Visits (engineer_id VARCHAR); CREATE TABLE Maintenance_Engineers (engineer_id VARCHAR, first_name VARCHAR, last_name VARCHAR) | SELECT T1.engineer_id, T1.first_name, T1.last_name FROM Maintenance_Engineers AS T1 JOIN Engineer_Visits AS T2 ON T1.engineer_id = T2.engineer_id GROUP BY T1.engineer_id ORDER BY COUNT(*) LIMIT 1 | ### Context: CREATE TABLE Engineer_Visits (engineer_id VARCHAR); CREATE TABLE Maintenance_Engineers (engineer_id VARCHAR, first_name VARCHAR, last_name VARCHAR) ### Question: Among those engineers who have visited, which engineer makes the least number of visits? List the engineer id, first name and last name. ### Answer: SELECT T1.engineer_id, T1.first_name, T1.last_name FROM Maintenance_Engineers AS T1 JOIN Engineer_Visits AS T2 ON T1.engineer_id = T2.engineer_id GROUP BY T1.engineer_id ORDER BY COUNT(*) LIMIT 1 |
Which staff have contacted which engineers? List the staff name and the engineer first name and last name. | CREATE TABLE Engineer_Visits (contact_staff_id VARCHAR, engineer_id VARCHAR); CREATE TABLE Staff (staff_name VARCHAR, staff_id VARCHAR); CREATE TABLE Maintenance_Engineers (first_name VARCHAR, last_name VARCHAR, engineer_id VARCHAR) | SELECT T1.staff_name, T3.first_name, T3.last_name FROM Staff AS T1 JOIN Engineer_Visits AS T2 ON T1.staff_id = T2.contact_staff_id JOIN Maintenance_Engineers AS T3 ON T2.engineer_id = T3.engineer_id | ### Context: CREATE TABLE Engineer_Visits (contact_staff_id VARCHAR, engineer_id VARCHAR); CREATE TABLE Staff (staff_name VARCHAR, staff_id VARCHAR); CREATE TABLE Maintenance_Engineers (first_name VARCHAR, last_name VARCHAR, engineer_id VARCHAR) ### Question: Which staff have contacted which engineers? List the staff name and the engineer first name and last name. ### Answer: SELECT T1.staff_name, T3.first_name, T3.last_name FROM Staff AS T1 JOIN Engineer_Visits AS T2 ON T1.staff_id = T2.contact_staff_id JOIN Maintenance_Engineers AS T3 ON T2.engineer_id = T3.engineer_id |
Which fault log included the most number of faulty parts? List the fault log id, description and record time. | CREATE TABLE Fault_Log (fault_log_entry_id VARCHAR, fault_description VARCHAR, fault_log_entry_datetime VARCHAR); CREATE TABLE Fault_Log_Parts (fault_log_entry_id VARCHAR) | SELECT T1.fault_log_entry_id, T1.fault_description, T1.fault_log_entry_datetime FROM Fault_Log AS T1 JOIN Fault_Log_Parts AS T2 ON T1.fault_log_entry_id = T2.fault_log_entry_id GROUP BY T1.fault_log_entry_id ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE Fault_Log (fault_log_entry_id VARCHAR, fault_description VARCHAR, fault_log_entry_datetime VARCHAR); CREATE TABLE Fault_Log_Parts (fault_log_entry_id VARCHAR) ### Question: Which fault log included the most number of faulty parts? List the fault log id, description and record time. ### Answer: SELECT T1.fault_log_entry_id, T1.fault_description, T1.fault_log_entry_datetime FROM Fault_Log AS T1 JOIN Fault_Log_Parts AS T2 ON T1.fault_log_entry_id = T2.fault_log_entry_id GROUP BY T1.fault_log_entry_id ORDER BY COUNT(*) DESC LIMIT 1 |
Which skill is used in fixing the most number of faults? List the skill id and description. | CREATE TABLE Skills (skill_id VARCHAR, skill_description VARCHAR); CREATE TABLE Skills_Required_To_Fix (skill_id VARCHAR) | SELECT T1.skill_id, T1.skill_description FROM Skills AS T1 JOIN Skills_Required_To_Fix AS T2 ON T1.skill_id = T2.skill_id GROUP BY T1.skill_id ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE Skills (skill_id VARCHAR, skill_description VARCHAR); CREATE TABLE Skills_Required_To_Fix (skill_id VARCHAR) ### Question: Which skill is used in fixing the most number of faults? List the skill id and description. ### Answer: SELECT T1.skill_id, T1.skill_description FROM Skills AS T1 JOIN Skills_Required_To_Fix AS T2 ON T1.skill_id = T2.skill_id GROUP BY T1.skill_id ORDER BY COUNT(*) DESC LIMIT 1 |
What are all the distinct asset models? | CREATE TABLE Assets (asset_model VARCHAR) | SELECT DISTINCT asset_model FROM Assets | ### Context: CREATE TABLE Assets (asset_model VARCHAR) ### Question: What are all the distinct asset models? ### Answer: SELECT DISTINCT asset_model FROM Assets |
List the all the assets make, model, details by the disposed date ascendingly. | CREATE TABLE Assets (asset_make VARCHAR, asset_model VARCHAR, asset_details VARCHAR, asset_disposed_date VARCHAR) | SELECT asset_make, asset_model, asset_details FROM Assets ORDER BY asset_disposed_date | ### Context: CREATE TABLE Assets (asset_make VARCHAR, asset_model VARCHAR, asset_details VARCHAR, asset_disposed_date VARCHAR) ### Question: List the all the assets make, model, details by the disposed date ascendingly. ### Answer: SELECT asset_make, asset_model, asset_details FROM Assets ORDER BY asset_disposed_date |
Which part has the least chargeable amount? List the part id and amount. | CREATE TABLE Parts (part_id VARCHAR, chargeable_amount VARCHAR) | SELECT part_id, chargeable_amount FROM Parts ORDER BY chargeable_amount LIMIT 1 | ### Context: CREATE TABLE Parts (part_id VARCHAR, chargeable_amount VARCHAR) ### Question: Which part has the least chargeable amount? List the part id and amount. ### Answer: SELECT part_id, chargeable_amount FROM Parts ORDER BY chargeable_amount LIMIT 1 |
Which company started the earliest the maintenance contract? Show the company name. | CREATE TABLE Third_Party_Companies (company_name VARCHAR, company_id VARCHAR); CREATE TABLE Maintenance_Contracts (maintenance_contract_company_id VARCHAR, contract_start_date VARCHAR) | SELECT T1.company_name FROM Third_Party_Companies AS T1 JOIN Maintenance_Contracts AS T2 ON T1.company_id = T2.maintenance_contract_company_id ORDER BY T2.contract_start_date LIMIT 1 | ### Context: CREATE TABLE Third_Party_Companies (company_name VARCHAR, company_id VARCHAR); CREATE TABLE Maintenance_Contracts (maintenance_contract_company_id VARCHAR, contract_start_date VARCHAR) ### Question: Which company started the earliest the maintenance contract? Show the company name. ### Answer: SELECT T1.company_name FROM Third_Party_Companies AS T1 JOIN Maintenance_Contracts AS T2 ON T1.company_id = T2.maintenance_contract_company_id ORDER BY T2.contract_start_date LIMIT 1 |
What is the description of the type of the company who concluded its contracts most recently? | CREATE TABLE Maintenance_Contracts (maintenance_contract_company_id VARCHAR, contract_end_date VARCHAR); CREATE TABLE Third_Party_Companies (company_name VARCHAR, company_id VARCHAR, company_type_code VARCHAR); CREATE TABLE Ref_Company_Types (company_type_code VARCHAR) | SELECT T1.company_name FROM Third_Party_Companies AS T1 JOIN Maintenance_Contracts AS T2 ON T1.company_id = T2.maintenance_contract_company_id JOIN Ref_Company_Types AS T3 ON T1.company_type_code = T3.company_type_code ORDER BY T2.contract_end_date DESC LIMIT 1 | ### Context: CREATE TABLE Maintenance_Contracts (maintenance_contract_company_id VARCHAR, contract_end_date VARCHAR); CREATE TABLE Third_Party_Companies (company_name VARCHAR, company_id VARCHAR, company_type_code VARCHAR); CREATE TABLE Ref_Company_Types (company_type_code VARCHAR) ### Question: What is the description of the type of the company who concluded its contracts most recently? ### Answer: SELECT T1.company_name FROM Third_Party_Companies AS T1 JOIN Maintenance_Contracts AS T2 ON T1.company_id = T2.maintenance_contract_company_id JOIN Ref_Company_Types AS T3 ON T1.company_type_code = T3.company_type_code ORDER BY T2.contract_end_date DESC LIMIT 1 |
Which gender makes up the majority of the staff? | CREATE TABLE staff (gender VARCHAR) | SELECT gender FROM staff GROUP BY gender ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE staff (gender VARCHAR) ### Question: Which gender makes up the majority of the staff? ### Answer: SELECT gender FROM staff GROUP BY gender ORDER BY COUNT(*) DESC LIMIT 1 |
How many engineers did each staff contact? List both the contact staff name and number of engineers contacted. | CREATE TABLE Engineer_Visits (contact_staff_id VARCHAR); CREATE TABLE Staff (staff_name VARCHAR, staff_id VARCHAR) | SELECT T1.staff_name, COUNT(*) FROM Staff AS T1 JOIN Engineer_Visits AS T2 ON T1.staff_id = T2.contact_staff_id GROUP BY T1.staff_name | ### Context: CREATE TABLE Engineer_Visits (contact_staff_id VARCHAR); CREATE TABLE Staff (staff_name VARCHAR, staff_id VARCHAR) ### Question: How many engineers did each staff contact? List both the contact staff name and number of engineers contacted. ### Answer: SELECT T1.staff_name, COUNT(*) FROM Staff AS T1 JOIN Engineer_Visits AS T2 ON T1.staff_id = T2.contact_staff_id GROUP BY T1.staff_name |
Which assets did not incur any fault log? List the asset model. | CREATE TABLE Fault_Log (asset_model VARCHAR, asset_id VARCHAR); CREATE TABLE Assets (asset_model VARCHAR, asset_id VARCHAR) | SELECT asset_model FROM Assets WHERE NOT asset_id IN (SELECT asset_id FROM Fault_Log) | ### Context: CREATE TABLE Fault_Log (asset_model VARCHAR, asset_id VARCHAR); CREATE TABLE Assets (asset_model VARCHAR, asset_id VARCHAR) ### Question: Which assets did not incur any fault log? List the asset model. ### Answer: SELECT asset_model FROM Assets WHERE NOT asset_id IN (SELECT asset_id FROM Fault_Log) |
list the local authorities and services provided by all stations. | CREATE TABLE station (local_authority VARCHAR, services VARCHAR) | SELECT local_authority, services FROM station | ### Context: CREATE TABLE station (local_authority VARCHAR, services VARCHAR) ### Question: list the local authorities and services provided by all stations. ### Answer: SELECT local_authority, services FROM station |
show all train numbers and names ordered by their time from early to late. | CREATE TABLE train (train_number VARCHAR, name VARCHAR, TIME VARCHAR) | SELECT train_number, name FROM train ORDER BY TIME | ### Context: CREATE TABLE train (train_number VARCHAR, name VARCHAR, TIME VARCHAR) ### Question: show all train numbers and names ordered by their time from early to late. ### Answer: SELECT train_number, name FROM train ORDER BY TIME |
Give me the times and numbers of all trains that go to Chennai, ordered by time. | CREATE TABLE train (TIME VARCHAR, train_number VARCHAR, destination VARCHAR) | SELECT TIME, train_number FROM train WHERE destination = 'Chennai' ORDER BY TIME | ### Context: CREATE TABLE train (TIME VARCHAR, train_number VARCHAR, destination VARCHAR) ### Question: Give me the times and numbers of all trains that go to Chennai, ordered by time. ### Answer: SELECT TIME, train_number FROM train WHERE destination = 'Chennai' ORDER BY TIME |
How many trains have 'Express' in their names? | CREATE TABLE train (name VARCHAR) | SELECT COUNT(*) FROM train WHERE name LIKE "%Express%" | ### Context: CREATE TABLE train (name VARCHAR) ### Question: How many trains have 'Express' in their names? ### Answer: SELECT COUNT(*) FROM train WHERE name LIKE "%Express%" |
Find the number and time of the train that goes from Chennai to Guruvayur. | CREATE TABLE train (train_number VARCHAR, TIME VARCHAR, origin VARCHAR, destination VARCHAR) | SELECT train_number, TIME FROM train WHERE origin = 'Chennai' AND destination = 'Guruvayur' | ### Context: CREATE TABLE train (train_number VARCHAR, TIME VARCHAR, origin VARCHAR, destination VARCHAR) ### Question: Find the number and time of the train that goes from Chennai to Guruvayur. ### Answer: SELECT train_number, TIME FROM train WHERE origin = 'Chennai' AND destination = 'Guruvayur' |
Find the number of trains starting from each origin. | CREATE TABLE train (origin VARCHAR) | SELECT origin, COUNT(*) FROM train GROUP BY origin | ### Context: CREATE TABLE train (origin VARCHAR) ### Question: Find the number of trains starting from each origin. ### Answer: SELECT origin, COUNT(*) FROM train GROUP BY origin |
Find the name of the train whose route runs through greatest number of stations. | CREATE TABLE route (train_id VARCHAR); CREATE TABLE train (name VARCHAR, id VARCHAR) | SELECT t1.name FROM train AS t1 JOIN route AS t2 ON t1.id = t2.train_id GROUP BY t2.train_id ORDER BY COUNT(*) DESC LIMIT 1 | ### Context: CREATE TABLE route (train_id VARCHAR); CREATE TABLE train (name VARCHAR, id VARCHAR) ### Question: Find the name of the train whose route runs through greatest number of stations. ### Answer: SELECT t1.name FROM train AS t1 JOIN route AS t2 ON t1.id = t2.train_id GROUP BY t2.train_id ORDER BY COUNT(*) DESC LIMIT 1 |
Find the number of trains for each station, as well as the station network name and services. | CREATE TABLE route (station_id VARCHAR); CREATE TABLE station (network_name VARCHAR, services VARCHAR, id VARCHAR) | SELECT COUNT(*), t1.network_name, t1.services FROM station AS t1 JOIN route AS t2 ON t1.id = t2.station_id GROUP BY t2.station_id | ### Context: CREATE TABLE route (station_id VARCHAR); CREATE TABLE station (network_name VARCHAR, services VARCHAR, id VARCHAR) ### Question: Find the number of trains for each station, as well as the station network name and services. ### Answer: SELECT COUNT(*), t1.network_name, t1.services FROM station AS t1 JOIN route AS t2 ON t1.id = t2.station_id GROUP BY t2.station_id |
What is the average high temperature for each day of week? | CREATE TABLE weekly_weather (day_of_week VARCHAR, high_temperature INTEGER) | SELECT AVG(high_temperature), day_of_week FROM weekly_weather GROUP BY day_of_week | ### Context: CREATE TABLE weekly_weather (day_of_week VARCHAR, high_temperature INTEGER) ### Question: What is the average high temperature for each day of week? ### Answer: SELECT AVG(high_temperature), day_of_week FROM weekly_weather GROUP BY day_of_week |
Give me the maximum low temperature and average precipitation at the Amersham station. | CREATE TABLE weekly_weather (low_temperature INTEGER, precipitation INTEGER, station_id VARCHAR); CREATE TABLE station (id VARCHAR, network_name VARCHAR) | SELECT MAX(t1.low_temperature), AVG(t1.precipitation) FROM weekly_weather AS t1 JOIN station AS t2 ON t1.station_id = t2.id WHERE t2.network_name = "Amersham" | ### Context: CREATE TABLE weekly_weather (low_temperature INTEGER, precipitation INTEGER, station_id VARCHAR); CREATE TABLE station (id VARCHAR, network_name VARCHAR) ### Question: Give me the maximum low temperature and average precipitation at the Amersham station. ### Answer: SELECT MAX(t1.low_temperature), AVG(t1.precipitation) FROM weekly_weather AS t1 JOIN station AS t2 ON t1.station_id = t2.id WHERE t2.network_name = "Amersham" |
Find names and times of trains that run through stations for the local authority Chiltern. | CREATE TABLE station (id VARCHAR, local_authority VARCHAR); CREATE TABLE route (station_id VARCHAR, train_id VARCHAR); CREATE TABLE train (name VARCHAR, time VARCHAR, id VARCHAR) | SELECT t3.name, t3.time FROM station AS t1 JOIN route AS t2 ON t1.id = t2.station_id JOIN train AS t3 ON t2.train_id = t3.id WHERE t1.local_authority = "Chiltern" | ### Context: CREATE TABLE station (id VARCHAR, local_authority VARCHAR); CREATE TABLE route (station_id VARCHAR, train_id VARCHAR); CREATE TABLE train (name VARCHAR, time VARCHAR, id VARCHAR) ### Question: Find names and times of trains that run through stations for the local authority Chiltern. ### Answer: SELECT t3.name, t3.time FROM station AS t1 JOIN route AS t2 ON t1.id = t2.station_id JOIN train AS t3 ON t2.train_id = t3.id WHERE t1.local_authority = "Chiltern" |
How many different services are provided by all stations? | CREATE TABLE station (services VARCHAR) | SELECT COUNT(DISTINCT services) FROM station | ### Context: CREATE TABLE station (services VARCHAR) ### Question: How many different services are provided by all stations? ### Answer: SELECT COUNT(DISTINCT services) FROM station |
Find the id and local authority of the station with has the highest average high temperature. | CREATE TABLE weekly_weather (station_id VARCHAR); CREATE TABLE station (id VARCHAR, local_authority VARCHAR) | SELECT t2.id, t2.local_authority FROM weekly_weather AS t1 JOIN station AS t2 ON t1.station_id = t2.id GROUP BY t1.station_id ORDER BY AVG(high_temperature) DESC LIMIT 1 | ### Context: CREATE TABLE weekly_weather (station_id VARCHAR); CREATE TABLE station (id VARCHAR, local_authority VARCHAR) ### Question: Find the id and local authority of the station with has the highest average high temperature. ### Answer: SELECT t2.id, t2.local_authority FROM weekly_weather AS t1 JOIN station AS t2 ON t1.station_id = t2.id GROUP BY t1.station_id ORDER BY AVG(high_temperature) DESC LIMIT 1 |
Find the id and local authority of the station whose maximum precipitation is higher than 50. | CREATE TABLE weekly_weather (station_id VARCHAR, precipitation INTEGER); CREATE TABLE station (id VARCHAR, local_authority VARCHAR) | SELECT t2.id, t2.local_authority FROM weekly_weather AS t1 JOIN station AS t2 ON t1.station_id = t2.id GROUP BY t1.station_id HAVING MAX(t1.precipitation) > 50 | ### Context: CREATE TABLE weekly_weather (station_id VARCHAR, precipitation INTEGER); CREATE TABLE station (id VARCHAR, local_authority VARCHAR) ### Question: Find the id and local authority of the station whose maximum precipitation is higher than 50. ### Answer: SELECT t2.id, t2.local_authority FROM weekly_weather AS t1 JOIN station AS t2 ON t1.station_id = t2.id GROUP BY t1.station_id HAVING MAX(t1.precipitation) > 50 |
show the lowest low temperature and highest wind speed in miles per hour. | CREATE TABLE weekly_weather (low_temperature INTEGER, wind_speed_mph INTEGER) | SELECT MIN(low_temperature), MAX(wind_speed_mph) FROM weekly_weather | ### Context: CREATE TABLE weekly_weather (low_temperature INTEGER, wind_speed_mph INTEGER) ### Question: show the lowest low temperature and highest wind speed in miles per hour. ### Answer: SELECT MIN(low_temperature), MAX(wind_speed_mph) FROM weekly_weather |
Find the origins from which more than 1 train starts. | CREATE TABLE train (origin VARCHAR) | SELECT origin FROM train GROUP BY origin HAVING COUNT(*) > 1 | ### Context: CREATE TABLE train (origin VARCHAR) ### Question: Find the origins from which more than 1 train starts. ### Answer: SELECT origin FROM train GROUP BY origin HAVING COUNT(*) > 1 |
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