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Question: Find all the name of documents without any sections. Context: CREATE TABLE document_sections (document_name VARCHAR, document_code VARCHAR); CREATE TABLE documents (document_name VARCHAR, document_code VARCHAR) Answer:
SELECT document_name FROM documents WHERE NOT document_code IN (SELECT document_code FROM document_sections)
Question: List all the username and passwords of users with the most popular role. Context: CREATE TABLE users (user_name VARCHAR, password VARCHAR, role_code VARCHAR) Answer:
SELECT user_name, password FROM users GROUP BY role_code ORDER BY COUNT(*) DESC LIMIT 1
Question: Find the average access counts of documents with functional area "Acknowledgement". Context: CREATE TABLE document_functional_areas (document_code VARCHAR, functional_area_code VARCHAR); CREATE TABLE documents (access_count INTEGER, document_code VARCHAR); CREATE TABLE functional_areas (functional_area_code VARCHAR, functional_area_description VARCHAR) Answer:
SELECT AVG(t1.access_count) FROM documents AS t1 JOIN document_functional_areas AS t2 ON t1.document_code = t2.document_code JOIN functional_areas AS t3 ON t2.functional_area_code = t3.functional_area_code WHERE t3.functional_area_description = "Acknowledgement"
Question: Find names of the document without any images. Context: CREATE TABLE document_sections_images (section_id VARCHAR); CREATE TABLE documents (document_name VARCHAR); CREATE TABLE documents (document_name VARCHAR, document_code VARCHAR); CREATE TABLE document_sections (document_code VARCHAR, section_id VARCHAR) Answer:
SELECT document_name FROM documents EXCEPT SELECT t1.document_name FROM documents AS t1 JOIN document_sections AS t2 ON t1.document_code = t2.document_code JOIN document_sections_images AS t3 ON t2.section_id = t3.section_id
Question: What is the name of the document with the most number of sections? Context: CREATE TABLE document_sections (document_code VARCHAR); CREATE TABLE documents (document_name VARCHAR, document_code VARCHAR) Answer:
SELECT t1.document_name FROM documents AS t1 JOIN document_sections AS t2 ON t1.document_code = t2.document_code GROUP BY t1.document_code ORDER BY COUNT(*) DESC LIMIT 1
Question: List all the document names which contains "CV". Context: CREATE TABLE documents (document_name VARCHAR) Answer:
SELECT document_name FROM documents WHERE document_name LIKE "%CV%"
Question: How many users are logged in? Context: CREATE TABLE users (user_login VARCHAR) Answer:
SELECT COUNT(*) FROM users WHERE user_login = 1
Question: Find the description of the most popular role among the users that have logged in. Context: CREATE TABLE users (role_description VARCHAR, role_code VARCHAR, user_login VARCHAR); CREATE TABLE ROLES (role_description VARCHAR, role_code VARCHAR, user_login VARCHAR) Answer:
SELECT role_description FROM ROLES WHERE role_code = (SELECT role_code FROM users WHERE user_login = 1 GROUP BY role_code ORDER BY COUNT(*) DESC LIMIT 1)
Question: Find the average access count of documents with the least popular structure. Context: CREATE TABLE documents (access_count INTEGER, document_structure_code VARCHAR) Answer:
SELECT AVG(access_count) FROM documents GROUP BY document_structure_code ORDER BY COUNT(*) LIMIT 1
Question: List all the image name and URLs in the order of their names. Context: CREATE TABLE images (image_name VARCHAR, image_url VARCHAR) Answer:
SELECT image_name, image_url FROM images ORDER BY image_name
Question: Find the number of users in each role. Context: CREATE TABLE users (role_code VARCHAR) Answer:
SELECT COUNT(*), role_code FROM users GROUP BY role_code
Question: What document types have more than 2 corresponding documents? Context: CREATE TABLE documents (document_type_code VARCHAR) Answer:
SELECT document_type_code FROM documents GROUP BY document_type_code HAVING COUNT(*) > 2
Question: How many companies are there? Context: CREATE TABLE Companies (Id VARCHAR) Answer:
SELECT COUNT(*) FROM Companies
Question: List the names of companies in descending order of market value. Context: CREATE TABLE Companies (name VARCHAR, Market_Value_billion VARCHAR) Answer:
SELECT name FROM Companies ORDER BY Market_Value_billion DESC
Question: What are the names of companies whose headquarters are not "USA"? Context: CREATE TABLE Companies (name VARCHAR, Headquarters VARCHAR) Answer:
SELECT name FROM Companies WHERE Headquarters <> 'USA'
Question: What are the name and assets of each company, sorted in ascending order of company name? Context: CREATE TABLE Companies (name VARCHAR, Assets_billion VARCHAR) Answer:
SELECT name, Assets_billion FROM Companies ORDER BY name
Question: What are the average profits of companies? Context: CREATE TABLE Companies (Profits_billion INTEGER) Answer:
SELECT AVG(Profits_billion) FROM Companies
Question: What are the maximum and minimum sales of the companies whose industries are not "Banking". Context: CREATE TABLE Companies (Sales_billion INTEGER, Industry VARCHAR) Answer:
SELECT MAX(Sales_billion), MIN(Sales_billion) FROM Companies WHERE Industry <> "Banking"
Question: How many different industries are the companies in? Context: CREATE TABLE Companies (Industry VARCHAR) Answer:
SELECT COUNT(DISTINCT Industry) FROM Companies
Question: List the names of buildings in descending order of building height. Context: CREATE TABLE buildings (name VARCHAR, Height VARCHAR) Answer:
SELECT name FROM buildings ORDER BY Height DESC
Question: Find the stories of the building with the largest height. Context: CREATE TABLE buildings (Stories VARCHAR, Height VARCHAR) Answer:
SELECT Stories FROM buildings ORDER BY Height DESC LIMIT 1
Question: List the name of a building along with the name of a company whose office is in the building. Context: CREATE TABLE buildings (name VARCHAR, id VARCHAR); CREATE TABLE Office_locations (building_id VARCHAR, company_id VARCHAR); CREATE TABLE Companies (name VARCHAR, id VARCHAR) Answer:
SELECT T3.name, T2.name FROM Office_locations AS T1 JOIN buildings AS T2 ON T1.building_id = T2.id JOIN Companies AS T3 ON T1.company_id = T3.id
Question: Show the names of the buildings that have more than one company offices. Context: CREATE TABLE buildings (name VARCHAR, id VARCHAR); CREATE TABLE Companies (id VARCHAR); CREATE TABLE Office_locations (building_id VARCHAR, company_id VARCHAR) Answer:
SELECT T2.name FROM Office_locations AS T1 JOIN buildings AS T2 ON T1.building_id = T2.id JOIN Companies AS T3 ON T1.company_id = T3.id GROUP BY T1.building_id HAVING COUNT(*) > 1
Question: Show the name of the building that has the most company offices. Context: CREATE TABLE buildings (name VARCHAR, id VARCHAR); CREATE TABLE Companies (id VARCHAR); CREATE TABLE Office_locations (building_id VARCHAR, company_id VARCHAR) Answer:
SELECT T2.name FROM Office_locations AS T1 JOIN buildings AS T2 ON T1.building_id = T2.id JOIN Companies AS T3 ON T1.company_id = T3.id GROUP BY T1.building_id ORDER BY COUNT(*) DESC LIMIT 1
Question: Please show the names of the buildings whose status is "on-hold", in ascending order of stories. Context: CREATE TABLE buildings (name VARCHAR, Status VARCHAR, Stories VARCHAR) Answer:
SELECT name FROM buildings WHERE Status = "on-hold" ORDER BY Stories
Question: Please show each industry and the corresponding number of companies in that industry. Context: CREATE TABLE Companies (Industry VARCHAR) Answer:
SELECT Industry, COUNT(*) FROM Companies GROUP BY Industry
Question: Please show the industries of companies in descending order of the number of companies. Context: CREATE TABLE Companies (Industry VARCHAR) Answer:
SELECT Industry FROM Companies GROUP BY Industry ORDER BY COUNT(*) DESC
Question: List the industry shared by the most companies. Context: CREATE TABLE Companies (Industry VARCHAR) Answer:
SELECT Industry FROM Companies GROUP BY Industry ORDER BY COUNT(*) DESC LIMIT 1
Question: List the names of buildings that have no company office. Context: CREATE TABLE buildings (name VARCHAR, id VARCHAR, building_id VARCHAR); CREATE TABLE Office_locations (name VARCHAR, id VARCHAR, building_id VARCHAR) Answer:
SELECT name FROM buildings WHERE NOT id IN (SELECT building_id FROM Office_locations)
Question: Show the industries shared by companies whose headquarters are "USA" and companies whose headquarters are "China". Context: CREATE TABLE Companies (Industry VARCHAR, Headquarters VARCHAR) Answer:
SELECT Industry FROM Companies WHERE Headquarters = "USA" INTERSECT SELECT Industry FROM Companies WHERE Headquarters = "China"
Question: Find the number of companies whose industry is "Banking" or "Conglomerate", Context: CREATE TABLE Companies (Industry VARCHAR) Answer:
SELECT COUNT(*) FROM Companies WHERE Industry = "Banking" OR Industry = "Conglomerate"
Question: Show the headquarters shared by more than two companies. Context: CREATE TABLE Companies (Headquarters VARCHAR) Answer:
SELECT Headquarters FROM Companies GROUP BY Headquarters HAVING COUNT(*) > 2
Question: How many products are there? Context: CREATE TABLE Products (Id VARCHAR) Answer:
SELECT COUNT(*) FROM Products
Question: List the name of products in ascending order of price. Context: CREATE TABLE Products (Product_Name VARCHAR, Product_Price VARCHAR) Answer:
SELECT Product_Name FROM Products ORDER BY Product_Price
Question: What are the names and type codes of products? Context: CREATE TABLE Products (Product_Name VARCHAR, Product_Type_Code VARCHAR) Answer:
SELECT Product_Name, Product_Type_Code FROM Products
Question: Show the prices of the products named "Dining" or "Trading Policy". Context: CREATE TABLE Products (Product_Price VARCHAR, Product_Name VARCHAR) Answer:
SELECT Product_Price FROM Products WHERE Product_Name = "Dining" OR Product_Name = "Trading Policy"
Question: What is the average price for products? Context: CREATE TABLE Products (Product_Price INTEGER) Answer:
SELECT AVG(Product_Price) FROM Products
Question: What is the name of the product with the highest price? Context: CREATE TABLE Products (Product_Name VARCHAR, Product_Price VARCHAR) Answer:
SELECT Product_Name FROM Products ORDER BY Product_Price DESC LIMIT 1
Question: Show different type codes of products and the number of products with each type code. Context: CREATE TABLE Products (Product_Type_Code VARCHAR) Answer:
SELECT Product_Type_Code, COUNT(*) FROM Products GROUP BY Product_Type_Code
Question: Show the most common type code across products. Context: CREATE TABLE Products (Product_Type_Code VARCHAR) Answer:
SELECT Product_Type_Code FROM Products GROUP BY Product_Type_Code ORDER BY COUNT(*) DESC LIMIT 1
Question: Show the product type codes that have at least two products. Context: CREATE TABLE Products (Product_Type_Code VARCHAR) Answer:
SELECT Product_Type_Code FROM Products GROUP BY Product_Type_Code HAVING COUNT(*) >= 2
Question: Show the product type codes that have both products with price higher than 4500 and products with price lower than 3000. Context: CREATE TABLE Products (Product_Type_Code VARCHAR, Product_Price INTEGER) Answer:
SELECT Product_Type_Code FROM Products WHERE Product_Price > 4500 INTERSECT SELECT Product_Type_Code FROM Products WHERE Product_Price < 3000
Question: Show the names of products and the number of events they are in. Context: CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR) Answer:
SELECT T1.Product_Name, COUNT(*) FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name
Question: Show the names of products and the number of events they are in, sorted by the number of events in descending order. Context: CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR) Answer:
SELECT T1.Product_Name, COUNT(*) FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name ORDER BY COUNT(*) DESC
Question: Show the names of products that are in at least two events. Context: CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR) Answer:
SELECT T1.Product_Name FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name HAVING COUNT(*) >= 2
Question: Show the names of products that are in at least two events in ascending alphabetical order of product name. Context: CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_ID VARCHAR) Answer:
SELECT T1.Product_Name FROM Products AS T1 JOIN Products_in_Events AS T2 ON T1.Product_ID = T2.Product_ID GROUP BY T1.Product_Name HAVING COUNT(*) >= 2 ORDER BY T1.Product_Name
Question: List the names of products that are not in any event. Context: CREATE TABLE Products (Product_Name VARCHAR, Product_ID VARCHAR); CREATE TABLE Products_in_Events (Product_Name VARCHAR, Product_ID VARCHAR) Answer:
SELECT Product_Name FROM Products WHERE NOT Product_ID IN (SELECT Product_ID FROM Products_in_Events)
Question: How many artworks are there? Context: CREATE TABLE artwork (Id VARCHAR) Answer:
SELECT COUNT(*) FROM artwork
Question: List the name of artworks in ascending alphabetical order. Context: CREATE TABLE artwork (Name VARCHAR) Answer:
SELECT Name FROM artwork ORDER BY Name
Question: List the name of artworks whose type is not "Program Talent Show". Context: CREATE TABLE artwork (Name VARCHAR, TYPE VARCHAR) Answer:
SELECT Name FROM artwork WHERE TYPE <> "Program Talent Show"
Question: What are the names and locations of festivals? Context: CREATE TABLE festival_detail (Festival_Name VARCHAR, LOCATION VARCHAR) Answer:
SELECT Festival_Name, LOCATION FROM festival_detail
Question: What are the names of the chairs of festivals, sorted in ascending order of the year held? Context: CREATE TABLE festival_detail (Chair_Name VARCHAR, YEAR VARCHAR) Answer:
SELECT Chair_Name FROM festival_detail ORDER BY YEAR
Question: What is the location of the festival with the largest number of audience? Context: CREATE TABLE festival_detail (LOCATION VARCHAR, Num_of_Audience VARCHAR) Answer:
SELECT LOCATION FROM festival_detail ORDER BY Num_of_Audience DESC LIMIT 1
Question: What are the names of festivals held in year 2007? Context: CREATE TABLE festival_detail (Festival_Name VARCHAR, YEAR VARCHAR) Answer:
SELECT Festival_Name FROM festival_detail WHERE YEAR = 2007
Question: What is the average number of audience for festivals? Context: CREATE TABLE festival_detail (Num_of_Audience INTEGER) Answer:
SELECT AVG(Num_of_Audience) FROM festival_detail
Question: Show the names of the three most recent festivals. Context: CREATE TABLE festival_detail (Festival_Name VARCHAR, YEAR VARCHAR) Answer:
SELECT Festival_Name FROM festival_detail ORDER BY YEAR DESC LIMIT 3
Question: For each nomination, show the name of the artwork and name of the festival where it is nominated. Context: CREATE TABLE artwork (Name VARCHAR, Artwork_ID VARCHAR); CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR) Answer:
SELECT T2.Name, T3.Festival_Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID
Question: Show distinct types of artworks that are nominated in festivals in 2007. Context: CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE festival_detail (Festival_ID VARCHAR, Year VARCHAR); CREATE TABLE artwork (Type VARCHAR, Artwork_ID VARCHAR) Answer:
SELECT DISTINCT T2.Type FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID WHERE T3.Year = 2007
Question: Show the names of artworks in ascending order of the year they are nominated in. Context: CREATE TABLE artwork (Name VARCHAR, Artwork_ID VARCHAR); CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE festival_detail (Festival_ID VARCHAR, Year VARCHAR) Answer:
SELECT T2.Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID ORDER BY T3.Year
Question: Show the names of festivals that have nominated artworks of type "Program Talent Show". Context: CREATE TABLE nomination (Artwork_ID VARCHAR, Festival_ID VARCHAR); CREATE TABLE artwork (Artwork_ID VARCHAR, Type VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR) Answer:
SELECT T3.Festival_Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID WHERE T2.Type = "Program Talent Show"
Question: Show the ids and names of festivals that have at least two nominations for artworks. Context: CREATE TABLE nomination (Festival_ID VARCHAR, Artwork_ID VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR); CREATE TABLE artwork (Artwork_ID VARCHAR) Answer:
SELECT T1.Festival_ID, T3.Festival_Name FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID GROUP BY T1.Festival_ID HAVING COUNT(*) >= 2
Question: Show the id, name of each festival and the number of artworks it has nominated. Context: CREATE TABLE nomination (Festival_ID VARCHAR, Artwork_ID VARCHAR); CREATE TABLE festival_detail (Festival_Name VARCHAR, Festival_ID VARCHAR); CREATE TABLE artwork (Artwork_ID VARCHAR) Answer:
SELECT T1.Festival_ID, T3.Festival_Name, COUNT(*) FROM nomination AS T1 JOIN artwork AS T2 ON T1.Artwork_ID = T2.Artwork_ID JOIN festival_detail AS T3 ON T1.Festival_ID = T3.Festival_ID GROUP BY T1.Festival_ID
Question: Please show different types of artworks with the corresponding number of artworks of each type. Context: CREATE TABLE artwork (TYPE VARCHAR) Answer:
SELECT TYPE, COUNT(*) FROM artwork GROUP BY TYPE
Question: List the most common type of artworks. Context: CREATE TABLE artwork (TYPE VARCHAR) Answer:
SELECT TYPE FROM artwork GROUP BY TYPE ORDER BY COUNT(*) DESC LIMIT 1
Question: List the year in which there are more than one festivals. Context: CREATE TABLE festival_detail (YEAR VARCHAR) Answer:
SELECT YEAR FROM festival_detail GROUP BY YEAR HAVING COUNT(*) > 1
Question: List the name of artworks that are not nominated. Context: CREATE TABLE nomination (Name VARCHAR, Artwork_ID VARCHAR); CREATE TABLE Artwork (Name VARCHAR, Artwork_ID VARCHAR) Answer:
SELECT Name FROM Artwork WHERE NOT Artwork_ID IN (SELECT Artwork_ID FROM nomination)
Question: Show the number of audience in year 2008 or 2010. Context: CREATE TABLE festival_detail (Num_of_Audience VARCHAR, YEAR VARCHAR) Answer:
SELECT Num_of_Audience FROM festival_detail WHERE YEAR = 2008 OR YEAR = 2010
Question: What are the total number of the audiences who visited any of the festivals? Context: CREATE TABLE festival_detail (Num_of_Audience INTEGER) Answer:
SELECT SUM(Num_of_Audience) FROM festival_detail
Question: In which year are there festivals both inside the 'United States' and outside the 'United States'? Context: CREATE TABLE festival_detail (YEAR VARCHAR, LOCATION VARCHAR) Answer:
SELECT YEAR FROM festival_detail WHERE LOCATION = 'United States' INTERSECT SELECT YEAR FROM festival_detail WHERE LOCATION <> 'United States'
Question: How many premises are there? Context: CREATE TABLE premises (Id VARCHAR) Answer:
SELECT COUNT(*) FROM premises
Question: What are all the distinct premise types? Context: CREATE TABLE premises (premises_type VARCHAR) Answer:
SELECT DISTINCT premises_type FROM premises
Question: Find the types and details for all premises and order by the premise type. Context: CREATE TABLE premises (premises_type VARCHAR, premise_details VARCHAR) Answer:
SELECT premises_type, premise_details FROM premises ORDER BY premises_type
Question: Show each premise type and the number of premises in that type. Context: CREATE TABLE premises (premises_type VARCHAR) Answer:
SELECT premises_type, COUNT(*) FROM premises GROUP BY premises_type
Question: Show all distinct product categories along with the number of mailshots in each category. Context: CREATE TABLE mailshot_campaigns (product_category VARCHAR) Answer:
SELECT product_category, COUNT(*) FROM mailshot_campaigns GROUP BY product_category
Question: Show the name and phone of the customer without any mailshot. Context: CREATE TABLE customers (customer_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR) Answer:
SELECT customer_name, customer_phone FROM customers WHERE NOT customer_id IN (SELECT customer_id FROM mailshot_customers)
Question: Show the name and phone for customers with a mailshot with outcome code 'No Response'. Context: CREATE TABLE customers (customer_name VARCHAR, customer_phone VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR, outcome_code VARCHAR) Answer:
SELECT T1.customer_name, T1.customer_phone FROM customers AS T1 JOIN mailshot_customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.outcome_code = 'No Response'
Question: Show the outcome code of mailshots along with the number of mailshots in each outcome code. Context: CREATE TABLE mailshot_customers (outcome_code VARCHAR) Answer:
SELECT outcome_code, COUNT(*) FROM mailshot_customers GROUP BY outcome_code
Question: Show the names of customers who have at least 2 mailshots with outcome code 'Order'. Context: CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR) Answer:
SELECT T2.customer_name FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id WHERE outcome_code = 'Order' GROUP BY T1.customer_id HAVING COUNT(*) >= 2
Question: Show the names of customers who have the most mailshots. Context: CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR) Answer:
SELECT T2.customer_name FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) DESC LIMIT 1
Question: What are the name and payment method of customers who have both mailshots in 'Order' outcome and mailshots in 'No Response' outcome. Context: CREATE TABLE customers (customer_name VARCHAR, payment_method VARCHAR, customer_id VARCHAR); CREATE TABLE mailshot_customers (customer_id VARCHAR, outcome_code VARCHAR) Answer:
SELECT T2.customer_name, T2.payment_method FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.outcome_code = 'Order' INTERSECT SELECT T2.customer_name, T2.payment_method FROM mailshot_customers AS T1 JOIN customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.outcome_code = 'No Response'
Question: Show the premise type and address type code for all customer addresses. Context: CREATE TABLE premises (premises_type VARCHAR, premise_id VARCHAR); CREATE TABLE customer_addresses (address_type_code VARCHAR, premise_id VARCHAR) Answer:
SELECT T2.premises_type, T1.address_type_code FROM customer_addresses AS T1 JOIN premises AS T2 ON T1.premise_id = T2.premise_id
Question: What are the distinct address type codes for all customer addresses? Context: CREATE TABLE customer_addresses (address_type_code VARCHAR) Answer:
SELECT DISTINCT address_type_code FROM customer_addresses
Question: Show the shipping charge and customer id for customer orders with order status Cancelled or Paid. Context: CREATE TABLE customer_orders (order_shipping_charges VARCHAR, customer_id VARCHAR, order_status_code VARCHAR) Answer:
SELECT order_shipping_charges, customer_id FROM customer_orders WHERE order_status_code = 'Cancelled' OR order_status_code = 'Paid'
Question: Show the names of customers having an order with shipping method FedEx and order status Paid. Context: CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE customer_orders (customer_id VARCHAR) Answer:
SELECT T1.customer_name FROM customers AS T1 JOIN customer_orders AS T2 ON T1.customer_id = T2.customer_id WHERE shipping_method_code = 'FedEx' AND order_status_code = 'Paid'
Question: How many courses are there in total? Context: CREATE TABLE COURSE (Id VARCHAR) Answer:
SELECT COUNT(*) FROM COURSE
Question: How many courses have more than 2 credits? Context: CREATE TABLE COURSE (Credits INTEGER) Answer:
SELECT COUNT(*) FROM COURSE WHERE Credits > 2
Question: List all names of courses with 1 credit? Context: CREATE TABLE COURSE (CName VARCHAR, Credits VARCHAR) Answer:
SELECT CName FROM COURSE WHERE Credits = 1
Question: Which courses are taught on days MTW? Context: CREATE TABLE COURSE (CName VARCHAR, Days VARCHAR) Answer:
SELECT CName FROM COURSE WHERE Days = "MTW"
Question: What is the number of departments in Division "AS"? Context: CREATE TABLE DEPARTMENT (Division VARCHAR) Answer:
SELECT COUNT(*) FROM DEPARTMENT WHERE Division = "AS"
Question: What are the phones of departments in Room 268? Context: CREATE TABLE DEPARTMENT (DPhone VARCHAR, Room VARCHAR) Answer:
SELECT DPhone FROM DEPARTMENT WHERE Room = 268
Question: Find the number of students that have at least one grade "B". Context: CREATE TABLE ENROLLED_IN (StuID VARCHAR, Grade VARCHAR) Answer:
SELECT COUNT(DISTINCT StuID) FROM ENROLLED_IN WHERE Grade = "B"
Question: Find the max and min grade point for all letter grade. Context: CREATE TABLE GRADECONVERSION (gradepoint INTEGER) Answer:
SELECT MAX(gradepoint), MIN(gradepoint) FROM GRADECONVERSION
Question: Find the first names of students whose first names contain letter "a". Context: CREATE TABLE STUDENT (Fname VARCHAR) Answer:
SELECT DISTINCT Fname FROM STUDENT WHERE Fname LIKE '%a%'
Question: Find the first names and last names of male (sex is M) faculties who live in building NEB. Context: CREATE TABLE FACULTY (Fname VARCHAR, Lname VARCHAR, sex VARCHAR, Building VARCHAR) Answer:
SELECT Fname, Lname FROM FACULTY WHERE sex = "M" AND Building = "NEB"
Question: Find the rooms of faculties with rank professor who live in building NEB. Context: CREATE TABLE FACULTY (Room VARCHAR, Rank VARCHAR, Building VARCHAR) Answer:
SELECT Room FROM FACULTY WHERE Rank = "Professor" AND Building = "NEB"
Question: Find the department name that is in Building "Mergenthaler". Context: CREATE TABLE DEPARTMENT (DName VARCHAR, Building VARCHAR) Answer:
SELECT DName FROM DEPARTMENT WHERE Building = "Mergenthaler"
Question: List all information about courses sorted by credits in the ascending order. Context: CREATE TABLE COURSE (Credits VARCHAR) Answer:
SELECT * FROM COURSE ORDER BY Credits
Question: List the course name of courses sorted by credits. Context: CREATE TABLE COURSE (CName VARCHAR, Credits VARCHAR) Answer:
SELECT CName FROM COURSE ORDER BY Credits
Question: Find the first name of students in the descending order of age. Context: CREATE TABLE STUDENT (Fname VARCHAR, Age VARCHAR) Answer:
SELECT Fname FROM STUDENT ORDER BY Age DESC
Question: Find the last name of female (sex is F) students in the descending order of age. Context: CREATE TABLE STUDENT (LName VARCHAR, Sex VARCHAR, Age VARCHAR) Answer:
SELECT LName FROM STUDENT WHERE Sex = "F" ORDER BY Age DESC