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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?
|
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