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800 | SELECT T1.name FROM instructor AS T1 JOIN advisor AS T2 ON T1.id = T2.i_id GROUP BY T2.i_id HAVING COUNT(*) > 1 | Find the name of instructors who are advising more than one student. | CREATE TABLE advisor (i_id VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR) | Encontre o nome dos instrutores que estão aconselhando mais de um aluno. |
801 | SELECT T1.name FROM student AS T1 JOIN advisor AS T2 ON T1.id = T2.s_id GROUP BY T2.s_id HAVING COUNT(*) > 1 | Find the name of the students who have more than one advisor? | CREATE TABLE student (name VARCHAR, id VARCHAR); CREATE TABLE advisor (s_id VARCHAR) | Encontre o nome dos alunos que têm mais de um orientador? |
802 | SELECT COUNT(*), building FROM classroom WHERE capacity > 50 GROUP BY building | Find the number of rooms with more than 50 capacity for each building. | CREATE TABLE classroom (building VARCHAR, capacity INTEGER) | Encontre o número de quartos com mais de 50 capacidade para cada edifício. |
803 | SELECT MAX(capacity), AVG(capacity), building FROM classroom GROUP BY building | Find the maximum and average capacity among rooms in each building. | CREATE TABLE classroom (building VARCHAR, capacity INTEGER) | Encontre a capacidade máxima e média entre os quartos de cada edifício. |
804 | SELECT title FROM course GROUP BY title HAVING COUNT(*) > 1 | Find the title of the course that is offered by more than one department. | CREATE TABLE course (title VARCHAR) | Encontre o título do curso que é oferecido por mais de um departamento. |
805 | SELECT SUM(credits), dept_name FROM course GROUP BY dept_name | Find the total credits of courses provided by different department. | CREATE TABLE course (dept_name VARCHAR, credits INTEGER) | Encontre o total de créditos de cursos fornecidos por diferentes departamentos. |
806 | SELECT MIN(salary), dept_name FROM instructor GROUP BY dept_name HAVING AVG(salary) > (SELECT AVG(salary) FROM instructor) | Find the minimum salary for the departments whose average salary is above the average payment of all instructors. | CREATE TABLE instructor (dept_name VARCHAR, salary INTEGER) | Encontre o salário mínimo para os departamentos cujo salário médio está acima do pagamento médio de todos os instrutores. |
807 | SELECT COUNT(*), semester, YEAR FROM SECTION GROUP BY semester, YEAR | Find the number of courses provided in each semester and year. | CREATE TABLE SECTION (semester VARCHAR, YEAR VARCHAR) | Encontre o número de cursos oferecidos em cada semestre e ano. |
808 | SELECT YEAR FROM SECTION GROUP BY YEAR ORDER BY COUNT(*) DESC LIMIT 1 | Find the year which offers the largest number of courses. | CREATE TABLE SECTION (YEAR VARCHAR) | Encontre o ano que oferece o maior número de cursos. |
809 | SELECT semester, YEAR FROM SECTION GROUP BY semester, YEAR ORDER BY COUNT(*) DESC LIMIT 1 | Find the year and semester when offers the largest number of courses. | CREATE TABLE SECTION (semester VARCHAR, YEAR VARCHAR) | Encontre o ano e o semestre em que oferece o maior número de cursos. |
810 | SELECT dept_name FROM student GROUP BY dept_name ORDER BY COUNT(*) DESC LIMIT 1 | Find the name of department has the highest amount of students? | CREATE TABLE student (dept_name VARCHAR) | Descobrir o nome do departamento tem a maior quantidade de alunos? |
811 | SELECT COUNT(*), dept_name FROM student GROUP BY dept_name | Find the total number of students in each department. | CREATE TABLE student (dept_name VARCHAR) | Encontre o número total de alunos em cada departamento. |
812 | SELECT semester, YEAR FROM takes GROUP BY semester, YEAR ORDER BY COUNT(*) LIMIT 1 | Find the semester and year which has the least number of student taking any class. | CREATE TABLE takes (semester VARCHAR, YEAR VARCHAR) | Encontre o semestre e o ano que tem o menor número de alunos fazendo qualquer aula. |
813 | SELECT i_id FROM advisor AS T1 JOIN student AS T2 ON T1.s_id = T2.id WHERE T2.dept_name = 'History' | What is the id of the instructor who advises of all students from History department? | CREATE TABLE advisor (s_id VARCHAR); CREATE TABLE student (id VARCHAR, dept_name VARCHAR) | Qual é o ID do instrutor que aconselha de todos os alunos do departamento de História? |
814 | SELECT T2.name, T2.salary FROM advisor AS T1 JOIN instructor AS T2 ON T1.i_id = T2.id JOIN student AS T3 ON T1.s_id = T3.id WHERE T3.dept_name = 'History' | Find the name and salary of the instructors who are advisors of any student from History department? | CREATE TABLE instructor (name VARCHAR, salary VARCHAR, id VARCHAR); CREATE TABLE advisor (i_id VARCHAR, s_id VARCHAR); CREATE TABLE student (id VARCHAR, dept_name VARCHAR) | Encontre o nome e o salário dos instrutores que são conselheiros de qualquer aluno do departamento de História? |
815 | SELECT course_id FROM course EXCEPT SELECT course_id FROM prereq | Find the id of the courses that do not have any prerequisite? | CREATE TABLE prereq (course_id VARCHAR); CREATE TABLE course (course_id VARCHAR) | Encontrar o ID dos cursos que não têm qualquer pré-requisito? |
816 | SELECT title FROM course WHERE course_id IN (SELECT T1.prereq_id FROM prereq AS T1 JOIN course AS T2 ON T1.course_id = T2.course_id WHERE T2.title = 'International Finance') | What is the title of the prerequisite class of International Finance course? | CREATE TABLE course (title VARCHAR, course_id VARCHAR); CREATE TABLE prereq (prereq_id VARCHAR, course_id VARCHAR); CREATE TABLE course (course_id VARCHAR, title VARCHAR) | Qual é o título da classe pré-requisito do curso de Finanças Internacionais? |
817 | SELECT title FROM course WHERE course_id IN (SELECT T1.course_id FROM prereq AS T1 JOIN course AS T2 ON T1.prereq_id = T2.course_id WHERE T2.title = 'Differential Geometry') | Find the title of course whose prerequisite is course Differential Geometry. | CREATE TABLE prereq (course_id VARCHAR, prereq_id VARCHAR); CREATE TABLE course (title VARCHAR, course_id VARCHAR); CREATE TABLE course (course_id VARCHAR, title VARCHAR) | Encontre o título do curso cujo pré-requisito é o curso Geometria Diferencial. |
818 | SELECT name FROM student WHERE id IN (SELECT id FROM takes WHERE semester = 'Fall' AND YEAR = 2003) | Find the names of students who have taken any course in the fall semester of year 2003. | CREATE TABLE student (name VARCHAR, id VARCHAR, semester VARCHAR, YEAR VARCHAR); CREATE TABLE takes (name VARCHAR, id VARCHAR, semester VARCHAR, YEAR VARCHAR) | Encontre os nomes dos alunos que fizeram qualquer curso no semestre de outono do ano 2003. |
819 | SELECT T1.title FROM course AS T1 JOIN SECTION AS T2 ON T1.course_id = T2.course_id WHERE building = 'Chandler' AND semester = 'Fall' AND YEAR = 2010 | What is the title of the course that was offered at building Chandler during the fall semester in the year of 2010? | CREATE TABLE course (title VARCHAR, course_id VARCHAR); CREATE TABLE SECTION (course_id VARCHAR) | Qual é o título do curso que foi oferecido na construção de Chandler durante o semestre de outono no ano de 2010? |
820 | SELECT T1.name FROM instructor AS T1 JOIN teaches AS T2 ON T1.id = T2.id JOIN course AS T3 ON T2.course_id = T3.course_id WHERE T3.title = 'C Programming' | Find the name of the instructors who taught C Programming course before. | CREATE TABLE teaches (id VARCHAR, course_id VARCHAR); CREATE TABLE course (course_id VARCHAR, title VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR) | Encontre o nome dos instrutores que ensinaram o curso de Programação C antes. |
821 | SELECT T2.name, T2.salary FROM advisor AS T1 JOIN instructor AS T2 ON T1.i_id = T2.id JOIN student AS T3 ON T1.s_id = T3.id WHERE T3.dept_name = 'Math' | Find the name and salary of instructors who are advisors of the students from the Math department. | CREATE TABLE instructor (name VARCHAR, salary VARCHAR, id VARCHAR); CREATE TABLE advisor (i_id VARCHAR, s_id VARCHAR); CREATE TABLE student (id VARCHAR, dept_name VARCHAR) | Encontre o nome e o salário dos instrutores que são conselheiros dos alunos do departamento de Matemática. |
822 | SELECT T2.name FROM advisor AS T1 JOIN instructor AS T2 ON T1.i_id = T2.id JOIN student AS T3 ON T1.s_id = T3.id WHERE T3.dept_name = 'Math' ORDER BY T3.tot_cred | Find the name of instructors who are advisors of the students from the Math department, and sort the results by students' total credit. | CREATE TABLE student (id VARCHAR, dept_name VARCHAR, tot_cred VARCHAR); CREATE TABLE advisor (i_id VARCHAR, s_id VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR) | Encontre o nome dos instrutores que são conselheiros dos alunos do departamento de Matemática e classifique os resultados pelo crédito total dos alunos. |
823 | SELECT title FROM course WHERE course_id IN (SELECT T1.prereq_id FROM prereq AS T1 JOIN course AS T2 ON T1.course_id = T2.course_id WHERE T2.title = 'Mobile Computing') | What is the course title of the prerequisite of course Mobile Computing? | CREATE TABLE course (title VARCHAR, course_id VARCHAR); CREATE TABLE prereq (prereq_id VARCHAR, course_id VARCHAR); CREATE TABLE course (course_id VARCHAR, title VARCHAR) | Qual é o título do curso do pré-requisito do curso Computação Móvel? |
824 | SELECT T2.name FROM advisor AS T1 JOIN instructor AS T2 ON T1.i_id = T2.id JOIN student AS T3 ON T1.s_id = T3.id ORDER BY T3.tot_cred DESC LIMIT 1 | Find the name of instructor who is the advisor of the student who has the highest number of total credits. | CREATE TABLE student (id VARCHAR, tot_cred VARCHAR); CREATE TABLE advisor (i_id VARCHAR, s_id VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR) | Encontre o nome do instrutor que é o conselheiro do aluno que tem o maior número de créditos totais. |
825 | SELECT name FROM instructor WHERE NOT id IN (SELECT id FROM teaches) | Find the name of instructors who didn't teach any courses? | CREATE TABLE teaches (name VARCHAR, id VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR) | Encontre o nome dos instrutores que não ministraram nenhum curso? |
826 | SELECT id FROM instructor EXCEPT SELECT id FROM teaches | Find the id of instructors who didn't teach any courses? | CREATE TABLE teaches (id VARCHAR); CREATE TABLE instructor (id VARCHAR) | Encontrar o id de instrutores que não ensinaram nenhum curso? |
827 | SELECT name FROM instructor WHERE NOT id IN (SELECT id FROM teaches WHERE semester = 'Spring') | Find the names of instructors who didn't each any courses in any Spring semester. | CREATE TABLE teaches (name VARCHAR, id VARCHAR, semester VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR, semester VARCHAR) | Encontre os nomes dos instrutores que não fizeram nenhum curso em qualquer semestre da primavera. |
828 | SELECT dept_name FROM instructor GROUP BY dept_name ORDER BY AVG(salary) DESC LIMIT 1 | Find the name of the department which has the highest average salary of professors. | CREATE TABLE instructor (dept_name VARCHAR, salary INTEGER) | Encontre o nome do departamento que tem o salário médio mais alto dos professores. |
829 | SELECT AVG(T1.salary), COUNT(*) FROM instructor AS T1 JOIN department AS T2 ON T1.dept_name = T2.dept_name ORDER BY T2.budget DESC LIMIT 1 | Find the number and averaged salary of all instructors who are in the department with the highest budget. | CREATE TABLE department (dept_name VARCHAR, budget VARCHAR); CREATE TABLE instructor (salary INTEGER, dept_name VARCHAR) | Encontre o número e o salário médio de todos os instrutores que estão no departamento com o orçamento mais alto. |
830 | SELECT T3.title, T3.credits FROM classroom AS T1 JOIN SECTION AS T2 ON T1.building = T2.building AND T1.room_number = T2.room_number JOIN course AS T3 ON T2.course_id = T3.course_id WHERE T1.capacity = (SELECT MAX(capacity) FROM classroom) | What is the title and credits of the course that is taught in the largest classroom (with the highest capacity)? | CREATE TABLE SECTION (course_id VARCHAR, building VARCHAR, room_number VARCHAR); CREATE TABLE course (title VARCHAR, credits VARCHAR, course_id VARCHAR); CREATE TABLE classroom (capacity INTEGER, building VARCHAR, room_number VARCHAR); CREATE TABLE classroom (capacity INTEGER) | Qual é o título e os créditos do curso que é ministrado na maior sala de aula (com a maior capacidade)? |
831 | SELECT name FROM student WHERE NOT id IN (SELECT T1.id FROM takes AS T1 JOIN course AS T2 ON T1.course_id = T2.course_id WHERE T2.dept_name = 'Biology') | Find the name of students who didn't take any course from Biology department. | CREATE TABLE student (name VARCHAR, id VARCHAR); CREATE TABLE course (course_id VARCHAR, dept_name VARCHAR); CREATE TABLE takes (id VARCHAR, course_id VARCHAR) | Encontre o nome dos alunos que não fizeram nenhum curso do departamento de Biologia. |
832 | SELECT COUNT(DISTINCT T2.id), COUNT(DISTINCT T3.id), T3.dept_name FROM department AS T1 JOIN student AS T2 ON T1.dept_name = T2.dept_name JOIN instructor AS T3 ON T1.dept_name = T3.dept_name GROUP BY T3.dept_name | Find the total number of students and total number of instructors for each department. | CREATE TABLE department (dept_name VARCHAR); CREATE TABLE student (id VARCHAR, dept_name VARCHAR); CREATE TABLE instructor (dept_name VARCHAR, id VARCHAR) | Encontre o número total de alunos e o número total de instrutores para cada departamento. |
833 | SELECT T1.name FROM student AS T1 JOIN takes AS T2 ON T1.id = T2.id WHERE T2.course_id IN (SELECT T4.prereq_id FROM course AS T3 JOIN prereq AS T4 ON T3.course_id = T4.course_id WHERE T3.title = 'International Finance') | Find the name of students who have taken the prerequisite course of the course with title International Finance. | CREATE TABLE student (name VARCHAR, id VARCHAR); CREATE TABLE course (course_id VARCHAR, title VARCHAR); CREATE TABLE prereq (prereq_id VARCHAR, course_id VARCHAR); CREATE TABLE takes (id VARCHAR, course_id VARCHAR) | Encontre o nome dos alunos que fizeram o curso pré-requisito do curso com o título Finanças Internacionais. |
834 | SELECT name, salary FROM instructor WHERE salary < (SELECT AVG(salary) FROM instructor WHERE dept_name = 'Physics') | Find the name and salary of instructors whose salary is below the average salary of the instructors in the Physics department. | CREATE TABLE instructor (name VARCHAR, salary INTEGER, dept_name VARCHAR) | Encontre o nome e o salário dos instrutores cujo salário esteja abaixo do salário médio dos instrutores do departamento de Física. |
835 | SELECT T3.name FROM course AS T1 JOIN takes AS T2 ON T1.course_id = T2.course_id JOIN student AS T3 ON T2.id = T3.id WHERE T1.dept_name = 'Statistics' | Find the name of students who took some course offered by Statistics department. | CREATE TABLE student (name VARCHAR, id VARCHAR); CREATE TABLE takes (course_id VARCHAR, id VARCHAR); CREATE TABLE course (course_id VARCHAR, dept_name VARCHAR) | Encontre o nome dos alunos que fizeram algum curso oferecido pelo departamento de Estatística. |
836 | SELECT T2.building, T2.room_number, T2.semester, T2.year FROM course AS T1 JOIN SECTION AS T2 ON T1.course_id = T2.course_id WHERE T1.dept_name = 'Psychology' ORDER BY T1.title | Find the building, room number, semester and year of all courses offered by Psychology department sorted by course titles. | CREATE TABLE SECTION (building VARCHAR, room_number VARCHAR, semester VARCHAR, year VARCHAR, course_id VARCHAR); CREATE TABLE course (course_id VARCHAR, dept_name VARCHAR, title VARCHAR) | Encontre o edifício, o número do quarto, o semestre e o ano de todos os cursos oferecidos pelo departamento de Psicologia, classificados por títulos do curso. |
837 | SELECT name FROM instructor WHERE dept_name = 'Comp. Sci.' | Find the names of all instructors in computer science department | CREATE TABLE instructor (name VARCHAR, dept_name VARCHAR) | Encontre os nomes de todos os instrutores no departamento de ciência da computação |
838 | SELECT name FROM instructor WHERE dept_name = 'Comp. Sci.' AND salary > 80000 | Find the names of all instructors in Comp. Sci. department with salary > 80000. | CREATE TABLE instructor (name VARCHAR, dept_name VARCHAR, salary VARCHAR) | Encontre os nomes de todos os instrutores no departamento de Comp. Sci. com salário > 80000. |
839 | SELECT name, course_id FROM instructor AS T1 JOIN teaches AS T2 ON T1.ID = T2.ID | Find the names of all instructors who have taught some course and the course_id. | CREATE TABLE instructor (ID VARCHAR); CREATE TABLE teaches (ID VARCHAR) | Encontre os nomes de todos os instrutores que ensinaram algum curso e o course_id. |
840 | SELECT name, course_id FROM instructor AS T1 JOIN teaches AS T2 ON T1.ID = T2.ID WHERE T1.dept_name = 'Art' | Find the names of all instructors in the Art department who have taught some course and the course_id. | CREATE TABLE instructor (ID VARCHAR, dept_name VARCHAR); CREATE TABLE teaches (ID VARCHAR) | Encontre os nomes de todos os instrutores do departamento de Arte que ensinaram algum curso e o course_id. |
841 | SELECT name FROM instructor WHERE name LIKE '%dar%' | Find the names of all instructors whose name includes the substring “dar”. | CREATE TABLE instructor (name VARCHAR) | Encontre os nomes de todos os instrutores cujo nome inclui a substring “dar”. |
842 | SELECT DISTINCT name FROM instructor ORDER BY name | List in alphabetic order the names of all distinct instructors. | CREATE TABLE instructor (name VARCHAR) | Listar em ordem alfabética os nomes de todos os instrutores distintos. |
843 | SELECT course_id FROM SECTION WHERE semester = 'Fall' AND YEAR = 2009 UNION SELECT course_id FROM SECTION WHERE semester = 'Spring' AND YEAR = 2010 | Find courses that ran in Fall 2009 or in Spring 2010. | CREATE TABLE SECTION (course_id VARCHAR, semester VARCHAR, YEAR VARCHAR) | Encontre cursos que foram executados no outono de 2009 ou na primavera de 2010. |
844 | SELECT course_id FROM SECTION WHERE semester = 'Fall' AND YEAR = 2009 INTERSECT SELECT course_id FROM SECTION WHERE semester = 'Spring' AND YEAR = 2010 | Find courses that ran in Fall 2009 and in Spring 2010. | CREATE TABLE SECTION (course_id VARCHAR, semester VARCHAR, YEAR VARCHAR) | Encontre cursos que foram executados no outono de 2009 e na primavera de 2010. |
845 | SELECT course_id FROM SECTION WHERE semester = 'Fall' AND YEAR = 2009 EXCEPT SELECT course_id FROM SECTION WHERE semester = 'Spring' AND YEAR = 2010 | Find courses that ran in Fall 2009 but not in Spring 2010. | CREATE TABLE SECTION (course_id VARCHAR, semester VARCHAR, YEAR VARCHAR) | Encontre cursos que foram executados no outono de 2009, mas não na primavera de 2010. |
846 | SELECT DISTINCT salary FROM instructor WHERE salary < (SELECT MAX(salary) FROM instructor) | Find the salaries of all distinct instructors that are less than the largest salary. | CREATE TABLE instructor (salary INTEGER) | Encontre os salários de todos os instrutores distintos que são menores do que o maior salário. |
847 | SELECT COUNT(DISTINCT ID) FROM teaches WHERE semester = 'Spring' AND YEAR = 2010 | Find the total number of instructors who teach a course in the Spring 2010 semester. | CREATE TABLE teaches (ID VARCHAR, semester VARCHAR, YEAR VARCHAR) | Encontre o número total de instrutores que ensinam um curso no semestre da primavera de 2010. |
848 | SELECT dept_name, AVG(salary) FROM instructor GROUP BY dept_name HAVING AVG(salary) > 42000 | Find the names and average salaries of all departments whose average salary is greater than 42000. | CREATE TABLE instructor (dept_name VARCHAR, salary INTEGER) | Encontre os nomes e salários médios de todos os departamentos cujo salário médio é superior a 42000. |
849 | SELECT name FROM instructor WHERE salary > (SELECT MIN(salary) FROM instructor WHERE dept_name = 'Biology') | Find names of instructors with salary greater than that of some (at least one) instructor in the Biology department. | CREATE TABLE instructor (name VARCHAR, salary INTEGER, dept_name VARCHAR) | Encontre nomes de instrutores com salário maior do que o de algum (pelo menos um) instrutor no departamento de Biologia. |
850 | SELECT name FROM instructor WHERE salary > (SELECT MAX(salary) FROM instructor WHERE dept_name = 'Biology') | Find the names of all instructors whose salary is greater than the salary of all instructors in the Biology department. | CREATE TABLE instructor (name VARCHAR, salary INTEGER, dept_name VARCHAR) | Encontre os nomes de todos os instrutores cujo salário é maior do que o salário de todos os instrutores do departamento de Biologia. |
851 | SELECT COUNT(*) FROM debate | How many debates are there? | CREATE TABLE debate (Id VARCHAR) | Quantos debates existem? |
852 | SELECT Venue FROM debate ORDER BY Num_of_Audience | List the venues of debates in ascending order of the number of audience. | CREATE TABLE debate (Venue VARCHAR, Num_of_Audience VARCHAR) | Liste os locais de debates em ordem crescente do número de audiências. |
853 | SELECT Date, Venue FROM debate | What are the date and venue of each debate? | CREATE TABLE debate (Date VARCHAR, Venue VARCHAR) | Qual é a data e o local de cada debate? |
854 | SELECT Date FROM debate WHERE Num_of_Audience > 150 | List the dates of debates with number of audience bigger than 150 | CREATE TABLE debate (Date VARCHAR, Num_of_Audience INTEGER) | Listar as datas dos debates com o número de público maior que 150 |
855 | SELECT Name FROM people WHERE Age = 35 OR Age = 36 | Show the names of people aged either 35 or 36. | CREATE TABLE people (Name VARCHAR, Age VARCHAR) | Mostre os nomes das pessoas com 35 ou 36 anos. |
856 | SELECT Party FROM people ORDER BY Age LIMIT 1 | What is the party of the youngest people? | CREATE TABLE people (Party VARCHAR, Age VARCHAR) | Qual é o partido dos mais jovens? |
857 | SELECT Party, COUNT(*) FROM people GROUP BY Party | Show different parties of people along with the number of people in each party. | CREATE TABLE people (Party VARCHAR) | Mostre diferentes grupos de pessoas, juntamente com o número de pessoas em cada grupo. |
858 | SELECT Party FROM people GROUP BY Party ORDER BY COUNT(*) DESC LIMIT 1 | Show the party that has the most people. | CREATE TABLE people (Party VARCHAR) | Mostre a festa que tem mais pessoas. |
859 | SELECT DISTINCT Venue FROM debate | Show the distinct venues of debates | CREATE TABLE debate (Venue VARCHAR) | Mostrar os locais distintos de debates |
860 | SELECT T3.Name, T2.Date, T2.Venue FROM debate_people AS T1 JOIN debate AS T2 ON T1.Debate_ID = T2.Debate_ID JOIN people AS T3 ON T1.Affirmative = T3.People_ID | Show the names of people, and dates and venues of debates they are on the affirmative side. | CREATE TABLE people (Name VARCHAR, People_ID VARCHAR); CREATE TABLE debate (Date VARCHAR, Venue VARCHAR, Debate_ID VARCHAR); CREATE TABLE debate_people (Debate_ID VARCHAR, Affirmative VARCHAR) | Mostrar os nomes das pessoas, datas e locais de debates que estão no lado afirmativo. |
861 | SELECT T3.Name, T2.Date, T2.Venue FROM debate_people AS T1 JOIN debate AS T2 ON T1.Debate_ID = T2.Debate_ID JOIN people AS T3 ON T1.Negative = T3.People_ID ORDER BY T3.Name | Show the names of people, and dates and venues of debates they are on the negative side, ordered in ascending alphabetical order of name. | CREATE TABLE people (Name VARCHAR, People_ID VARCHAR); CREATE TABLE debate (Date VARCHAR, Venue VARCHAR, Debate_ID VARCHAR); CREATE TABLE debate_people (Debate_ID VARCHAR, Negative VARCHAR) | Mostrar os nomes das pessoas, e datas e locais de debates que estão no lado negativo, ordenados em ordem alfabética ascendente de nome. |
862 | SELECT T3.Name FROM debate_people AS T1 JOIN debate AS T2 ON T1.Debate_ID = T2.Debate_ID JOIN people AS T3 ON T1.Affirmative = T3.People_ID WHERE T2.Num_of_Audience > 200 | Show the names of people that are on affirmative side of debates with number of audience bigger than 200. | CREATE TABLE people (Name VARCHAR, People_ID VARCHAR); CREATE TABLE debate (Debate_ID VARCHAR, Num_of_Audience INTEGER); CREATE TABLE debate_people (Debate_ID VARCHAR, Affirmative VARCHAR) | Mostre os nomes das pessoas que estão no lado afirmativo dos debates com número de público maior que 200. |
863 | SELECT T2.Name, COUNT(*) FROM debate_people AS T1 JOIN people AS T2 ON T1.Affirmative = T2.People_ID GROUP BY T2.Name | Show the names of people and the number of times they have been on the affirmative side of debates. | CREATE TABLE people (Name VARCHAR, People_ID VARCHAR); CREATE TABLE debate_people (Affirmative VARCHAR) | Mostre os nomes das pessoas e o número de vezes que elas estiveram no lado afirmativo dos debates. |
864 | SELECT T2.Name FROM debate_people AS T1 JOIN people AS T2 ON T1.Negative = T2.People_ID GROUP BY T2.Name HAVING COUNT(*) >= 2 | Show the names of people who have been on the negative side of debates at least twice. | CREATE TABLE debate_people (Negative VARCHAR); CREATE TABLE people (Name VARCHAR, People_ID VARCHAR) | Mostre os nomes das pessoas que estiveram no lado negativo dos debates pelo menos duas vezes. |
865 | SELECT Name FROM people WHERE NOT People_id IN (SELECT Affirmative FROM debate_people) | List the names of people that have not been on the affirmative side of debates. | CREATE TABLE debate_people (Name VARCHAR, People_id VARCHAR, Affirmative VARCHAR); CREATE TABLE people (Name VARCHAR, People_id VARCHAR, Affirmative VARCHAR) | Liste os nomes de pessoas que não estiveram no lado afirmativo dos debates. |
866 | SELECT customer_details FROM customers ORDER BY customer_details | List the names of all the customers in alphabetical order. | CREATE TABLE customers (customer_details VARCHAR) | Liste os nomes de todos os clientes em ordem alfabética. |
867 | SELECT policy_type_code FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t2.customer_details = "Dayana Robel" | Find all the policy type codes associated with the customer "Dayana Robel" | CREATE TABLE customers (customer_id VARCHAR, customer_details VARCHAR); CREATE TABLE policies (customer_id VARCHAR) | Encontre todos os códigos de tipo de política associados ao cliente "Dayana Robel" |
868 | SELECT policy_type_code FROM policies GROUP BY policy_type_code ORDER BY COUNT(*) DESC LIMIT 1 | Which type of policy is most frequently used? Give me the policy type code. | CREATE TABLE policies (policy_type_code VARCHAR) | Que tipo de política é mais usado? Dê-me o código do tipo de política. |
869 | SELECT policy_type_code FROM policies GROUP BY policy_type_code HAVING COUNT(*) > 2 | Find all the policy types that are used by more than 2 customers. | CREATE TABLE policies (policy_type_code VARCHAR) | Encontre todos os tipos de políticas que são usados por mais de 2 clientes. |
870 | SELECT SUM(amount_piad), AVG(amount_piad) FROM claim_headers | Find the total and average amount paid in claim headers. | CREATE TABLE claim_headers (amount_piad INTEGER) | Encontre o valor total e médio pago em cabeçalhos de reivindicação. |
871 | SELECT SUM(t1.amount_claimed) FROM claim_headers AS t1 JOIN claims_documents AS t2 ON t1.claim_header_id = t2.claim_id WHERE t2.created_date = (SELECT created_date FROM claims_documents ORDER BY created_date LIMIT 1) | Find the total amount claimed in the most recently created document. | CREATE TABLE claim_headers (amount_claimed INTEGER, claim_header_id VARCHAR); CREATE TABLE claims_documents (claim_id VARCHAR, created_date VARCHAR); CREATE TABLE claims_documents (created_date VARCHAR) | Encontre o valor total reivindicado no documento criado mais recentemente. |
872 | SELECT t3.customer_details FROM claim_headers AS t1 JOIN policies AS t2 ON t1.policy_id = t2.policy_id JOIN customers AS t3 ON t2.customer_id = t3.customer_id WHERE t1.amount_claimed = (SELECT MAX(amount_claimed) FROM claim_headers) | What is the name of the customer who has made the largest amount of claim in a single claim? | CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR); CREATE TABLE claim_headers (amount_claimed INTEGER); CREATE TABLE policies (policy_id VARCHAR, customer_id VARCHAR); CREATE TABLE claim_headers (policy_id VARCHAR, amount_claimed INTEGER) | Qual é o nome do cliente que fez a maior quantidade de reivindicação em uma única reivindicação? |
873 | SELECT t3.customer_details FROM claim_headers AS t1 JOIN policies AS t2 ON t1.policy_id = t2.policy_id JOIN customers AS t3 ON t2.customer_id = t3.customer_id WHERE t1.amount_piad = (SELECT MIN(amount_piad) FROM claim_headers) | What is the name of the customer who has made the minimum amount of payment in one claim? | CREATE TABLE claim_headers (amount_piad INTEGER); CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR); CREATE TABLE policies (policy_id VARCHAR, customer_id VARCHAR); CREATE TABLE claim_headers (policy_id VARCHAR, amount_piad INTEGER) | Qual é o nome do cliente que fez o valor mínimo de pagamento em uma reclamação? |
874 | SELECT customer_details FROM customers EXCEPT SELECT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id | Find the names of customers who have no policies associated. | CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR); CREATE TABLE customers (customer_details VARCHAR); CREATE TABLE policies (customer_id VARCHAR) | Encontre os nomes dos clientes que não têm políticas associadas. |
875 | SELECT COUNT(*) FROM claims_processing_stages | How many claim processing stages are there in total? | CREATE TABLE claims_processing_stages (Id VARCHAR) | Quantos estágios de processamento de sinistros existem no total? |
876 | SELECT t2.claim_status_name FROM claims_processing AS t1 JOIN claims_processing_stages AS t2 ON t1.claim_stage_id = t2.claim_stage_id GROUP BY t1.claim_stage_id ORDER BY COUNT(*) DESC LIMIT 1 | What is the name of the claim processing stage that most of the claims are on? | CREATE TABLE claims_processing (claim_stage_id VARCHAR); CREATE TABLE claims_processing_stages (claim_status_name VARCHAR, claim_stage_id VARCHAR) | Qual é o nome da fase de processamento de reclamações em que a maioria das reivindicações estão? |
877 | SELECT customer_details FROM customers WHERE customer_details LIKE "%Diana%" | Find the names of customers whose name contains "Diana". | CREATE TABLE customers (customer_details VARCHAR) | Encontre os nomes dos clientes cujo nome contém "Diana". |
878 | SELECT DISTINCT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t1.policy_type_code = "Deputy" | Find the names of the customers who have an deputy policy. | CREATE TABLE policies (customer_id VARCHAR, policy_type_code VARCHAR); CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR) | Encontre os nomes dos clientes que têm uma política adjunta. |
879 | SELECT DISTINCT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t1.policy_type_code = "Deputy" OR t1.policy_type_code = "Uniform" | Find the names of customers who either have an deputy policy or uniformed policy. | CREATE TABLE policies (customer_id VARCHAR, policy_type_code VARCHAR); CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR) | Encontre os nomes dos clientes que têm uma política de vice ou uma política uniformizada. |
880 | SELECT customer_details FROM customers UNION SELECT staff_details FROM staff | Find the names of all the customers and staff members. | CREATE TABLE staff (customer_details VARCHAR, staff_details VARCHAR); CREATE TABLE customers (customer_details VARCHAR, staff_details VARCHAR) | Encontre os nomes de todos os clientes e funcionários. |
881 | SELECT policy_type_code, COUNT(*) FROM policies GROUP BY policy_type_code | Find the number of records of each policy type and its type code. | CREATE TABLE policies (policy_type_code VARCHAR) | Encontre o número de registros de cada tipo de política e seu código de tipo. |
882 | SELECT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id GROUP BY t2.customer_details ORDER BY COUNT(*) DESC LIMIT 1 | Find the name of the customer that has been involved in the most policies. | CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR); CREATE TABLE policies (customer_id VARCHAR) | Encontre o nome do cliente que esteve envolvido na maioria das políticas. |
883 | SELECT claim_status_description FROM claims_processing_stages WHERE claim_status_name = "Open" | What is the description of the claim status "Open"? | CREATE TABLE claims_processing_stages (claim_status_description VARCHAR, claim_status_name VARCHAR) | Qual é a descrição do status de reivindicação "Open"? |
884 | SELECT COUNT(DISTINCT claim_outcome_code) FROM claims_processing | How many distinct claim outcome codes are there? | CREATE TABLE claims_processing (claim_outcome_code VARCHAR) | Quantos códigos de resultado de reivindicação distintos existem? |
885 | SELECT t2.customer_details FROM policies AS t1 JOIN customers AS t2 ON t1.customer_id = t2.customer_id WHERE t1.start_date = (SELECT MAX(start_date) FROM policies) | Which customer is associated with the latest policy? | CREATE TABLE customers (customer_details VARCHAR, customer_id VARCHAR); CREATE TABLE policies (start_date INTEGER); CREATE TABLE policies (customer_id VARCHAR, start_date INTEGER) | Qual cliente está associado à política mais recente? |
886 | SELECT account_id, date_account_opened, account_name, other_account_details FROM Accounts | Show the id, the date of account opened, the account name, and other account detail for all accounts. | CREATE TABLE Accounts (account_id VARCHAR, date_account_opened VARCHAR, account_name VARCHAR, other_account_details VARCHAR) | Mostre o ID, a data de abertura da conta, o nome da conta e outros detalhes da conta para todas as contas. |
887 | SELECT T1.account_id, T1.date_account_opened, T1.account_name, T1.other_account_details FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = 'Meaghan' | Show the id, the account name, and other account details for all accounts by the customer with first name 'Meaghan'. | CREATE TABLE Accounts (account_id VARCHAR, date_account_opened VARCHAR, account_name VARCHAR, other_account_details VARCHAR, customer_id VARCHAR); CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR) | Mostre o ID, o nome da conta e outros detalhes da conta para todas as contas pelo cliente com o primeiro nome 'Meaghan'. |
888 | SELECT T1.account_name, T1.other_account_details FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T2.customer_first_name = "Meaghan" AND T2.customer_last_name = "Keeling" | Show the account name and other account detail for all accounts by the customer with first name Meaghan and last name Keeling. | CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR, customer_last_name VARCHAR); CREATE TABLE Accounts (account_name VARCHAR, other_account_details VARCHAR, customer_id VARCHAR) | Mostre o nome da conta e outros detalhes da conta para todas as contas pelo cliente com o primeiro nome Meaghan e o último nome Keeling. |
889 | SELECT T2.customer_first_name, T2.customer_last_name FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE T1.account_name = "900" | Show the first name and last name for the customer with account name 900. | CREATE TABLE Accounts (customer_id VARCHAR, account_name VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR) | Mostrar o primeiro nome e o último nome do cliente com o nome da conta 900. |
890 | SELECT DISTINCT T1.customer_first_name, T1.customer_last_name, T1.phone_number FROM Customers AS T1 JOIN Accounts AS T2 ON T1.customer_id = T2.customer_id | Show the unique first names, last names, and phone numbers for all customers with any account. | CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, phone_number VARCHAR, customer_id VARCHAR) | Mostre os primeiros nomes, sobrenomes e números de telefone exclusivos para todos os clientes com qualquer conta. |
891 | SELECT customer_id FROM Customers EXCEPT SELECT customer_id FROM Accounts | Show customer ids who don't have an account. | CREATE TABLE Customers (customer_id VARCHAR); CREATE TABLE Accounts (customer_id VARCHAR) | Mostrar IDs de clientes que não têm uma conta. |
892 | SELECT COUNT(*), customer_id FROM Accounts GROUP BY customer_id | How many accounts does each customer have? List the number and customer id. | CREATE TABLE Accounts (customer_id VARCHAR) | Quantas contas cada cliente tem? Liste o número e o ID do cliente. |
893 | SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id ORDER BY COUNT(*) DESC LIMIT 1 | What is the customer id, first and last name with most number of accounts. | CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR) | Qual é o ID do cliente, primeiro e último nome com o maior número de contas? |
894 | SELECT T1.customer_id, T2.customer_first_name, T2.customer_last_name, COUNT(*) FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id | Show id, first name and last name for all customers and the number of accounts. | CREATE TABLE Accounts (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_last_name VARCHAR, customer_id VARCHAR) | Mostrar ID, primeiro nome e sobrenome para todos os clientes e o número de contas. |
895 | SELECT T2.customer_first_name, T1.customer_id FROM Accounts AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id GROUP BY T1.customer_id HAVING COUNT(*) >= 2 | Show first name and id for all customers with at least 2 accounts. | CREATE TABLE Customers (customer_first_name VARCHAR, customer_id VARCHAR); CREATE TABLE Accounts (customer_id VARCHAR) | Mostrar primeiro nome e ID para todos os clientes com pelo menos 2 contas. |
896 | SELECT gender, COUNT(*) FROM Customers GROUP BY gender | Show the number of customers for each gender. | CREATE TABLE Customers (gender VARCHAR) | Mostre o número de clientes para cada gênero. |
897 | SELECT COUNT(*) FROM Financial_transactions | How many transactions do we have? | CREATE TABLE Financial_transactions (Id VARCHAR) | Quantas transações temos? |
898 | SELECT COUNT(*), account_id FROM Financial_transactions | How many transaction does each account have? Show the number and account id. | CREATE TABLE Financial_transactions (account_id VARCHAR) | Quantas transações cada conta tem? Mostre o número e o ID da conta. |
899 | SELECT COUNT(*) FROM Financial_transactions AS T1 JOIN Accounts AS T2 ON T1.account_id = T2.account_id WHERE T2.account_name = "337" | How many transaction does account with name 337 have? | CREATE TABLE Accounts (account_id VARCHAR, account_name VARCHAR); CREATE TABLE Financial_transactions (account_id VARCHAR) | Quantas transações a conta com o nome 337 tem? |
Subsets and Splits