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SELECT T1.booking_start_date, T1.booking_start_date FROM Apartment_Bookings AS T1 JOIN Guests AS T2 ON T1.guest_id = T2.guest_id WHERE T2.gender_code = "Female"

Show the start dates and end dates of all the apartment bookings made by guests with gender code "Female".

CREATE TABLE Apartment_Bookings (booking_start_date VARCHAR, guest_id VARCHAR); CREATE TABLE Guests (guest_id VARCHAR, gender_code VARCHAR)

Mostrar as datas de início e término de todas as reservas de apartamentos feitas por hóspedes com o código de gênero "Fêmea".

SELECT T2.guest_first_name, T2.guest_last_name FROM Apartment_Bookings AS T1 JOIN Guests AS T2 ON T1.guest_id = T2.guest_id WHERE T1.booking_status_code = "Confirmed"

Show the first names and last names of all the guests that have apartment bookings with status code "Confirmed".

CREATE TABLE Apartment_Bookings (guest_id VARCHAR, booking_status_code VARCHAR); CREATE TABLE Guests (guest_first_name VARCHAR, guest_last_name VARCHAR, guest_id VARCHAR)

Mostre os primeiros nomes e sobrenomes de todos os hóspedes que tenham reservas de apartamentos com o código de status "Confirmado".

SELECT T1.facility_code FROM Apartment_Facilities AS T1 JOIN Apartments AS T2 ON T1.apt_id = T2.apt_id WHERE T2.bedroom_count > 4

Show the facility codes of apartments with more than 4 bedrooms.

CREATE TABLE Apartment_Facilities (facility_code VARCHAR, apt_id VARCHAR); CREATE TABLE Apartments (apt_id VARCHAR, bedroom_count INTEGER)

Mostrar os códigos de instalações de apartamentos com mais de 4 quartos.

SELECT SUM(T2.room_count) FROM Apartment_Facilities AS T1 JOIN Apartments AS T2 ON T1.apt_id = T2.apt_id WHERE T1.facility_code = "Gym"

Show the total number of rooms of all apartments with facility code "Gym".

CREATE TABLE Apartments (room_count INTEGER, apt_id VARCHAR); CREATE TABLE Apartment_Facilities (apt_id VARCHAR, facility_code VARCHAR)

Mostrar o número total de quartos de todos os apartamentos com o código de instalação "Gym".

SELECT SUM(T2.room_count) FROM Apartment_Buildings AS T1 JOIN Apartments AS T2 ON T1.building_id = T2.building_id WHERE T1.building_short_name = "Columbus Square"

Show the total number of rooms of the apartments in the building with short name "Columbus Square".

CREATE TABLE Apartment_Buildings (building_id VARCHAR, building_short_name VARCHAR); CREATE TABLE Apartments (room_count INTEGER, building_id VARCHAR)

Mostrar o número total de quartos dos apartamentos no edifício com o nome curto "Praça de Colombo".

SELECT T1.building_address FROM Apartment_Buildings AS T1 JOIN Apartments AS T2 ON T1.building_id = T2.building_id WHERE T2.bathroom_count > 2

Show the addresses of the buildings that have apartments with more than 2 bathrooms.

CREATE TABLE Apartment_Buildings (building_address VARCHAR, building_id VARCHAR); CREATE TABLE Apartments (building_id VARCHAR, bathroom_count INTEGER)

Mostrar os endereços dos edifícios que têm apartamentos com mais de 2 casas de banho.

SELECT T2.apt_type_code, T2.apt_number FROM Apartment_Buildings AS T1 JOIN Apartments AS T2 ON T1.building_id = T2.building_id WHERE T1.building_manager = "Kyle"

Show the apartment type codes and apartment numbers in the buildings managed by "Kyle".

CREATE TABLE Apartment_Buildings (building_id VARCHAR, building_manager VARCHAR); CREATE TABLE Apartments (apt_type_code VARCHAR, apt_number VARCHAR, building_id VARCHAR)

Mostrar os códigos de tipo de apartamento e números de apartamentos nos edifícios geridos por "Kyle".

SELECT booking_status_code, COUNT(*) FROM Apartment_Bookings GROUP BY booking_status_code

Show the booking status code and the corresponding number of bookings.

CREATE TABLE Apartment_Bookings (booking_status_code VARCHAR)

Mostrar o código de status da reserva e o número correspondente de reservas.

SELECT apt_number FROM Apartments ORDER BY room_count

Return all the apartment numbers sorted by the room count in ascending order.

CREATE TABLE Apartments (apt_number VARCHAR, room_count VARCHAR)

Devolva todos os números de apartamentos classificados pela contagem de quartos em ordem crescente.

SELECT apt_number FROM Apartments ORDER BY bedroom_count DESC LIMIT 1

Return the apartment number with the largest number of bedrooms.

CREATE TABLE Apartments (apt_number VARCHAR, bedroom_count VARCHAR)

Devolva o número do apartamento com o maior número de quartos.

SELECT apt_type_code, COUNT(*) FROM Apartments GROUP BY apt_type_code ORDER BY COUNT(*)

Show the apartment type codes and the corresponding number of apartments sorted by the number of apartments in ascending order.

CREATE TABLE Apartments (apt_type_code VARCHAR)

Mostrar os códigos de tipo de apartamento e o número correspondente de apartamentos classificados pelo número de apartamentos em ordem crescente.

SELECT apt_type_code FROM Apartments GROUP BY apt_type_code ORDER BY AVG(room_count) DESC LIMIT 3

Show the top 3 apartment type codes sorted by the average number of rooms in descending order.

CREATE TABLE Apartments (apt_type_code VARCHAR, room_count INTEGER)

Mostrar os 3 códigos de tipo de apartamento classificados pelo número médio de quartos em ordem decrescente.

SELECT apt_type_code, bathroom_count, bedroom_count FROM Apartments GROUP BY apt_type_code ORDER BY SUM(room_count) DESC LIMIT 1

Show the apartment type code that has the largest number of total rooms, together with the number of bathrooms and number of bedrooms.

CREATE TABLE Apartments (apt_type_code VARCHAR, bathroom_count VARCHAR, bedroom_count VARCHAR, room_count INTEGER)

Mostrar o código do tipo de apartamento que tem o maior número de quartos totais, juntamente com o número de casas de banho e número de quartos.

SELECT apt_type_code FROM Apartments GROUP BY apt_type_code ORDER BY COUNT(*) DESC LIMIT 1

Show the most common apartment type code.

CREATE TABLE Apartments (apt_type_code VARCHAR)

Mostre o código de tipo de apartamento mais comum.

SELECT apt_type_code FROM Apartments WHERE bathroom_count > 1 GROUP BY apt_type_code ORDER BY COUNT(*) DESC LIMIT 1

Show the most common apartment type code among apartments with more than 1 bathroom.

CREATE TABLE Apartments (apt_type_code VARCHAR, bathroom_count INTEGER)

Mostrar o código de tipo de apartamento mais comum entre os apartamentos com mais de 1 casa de banho.

SELECT apt_type_code, MAX(room_count), MIN(room_count) FROM Apartments GROUP BY apt_type_code

Show each apartment type code, and the maximum and minimum number of rooms for each type.

CREATE TABLE Apartments (apt_type_code VARCHAR, room_count INTEGER)

Mostre cada código de tipo de apartamento e o número máximo e mínimo de quartos para cada tipo.

SELECT gender_code, COUNT(*) FROM Guests GROUP BY gender_code ORDER BY COUNT(*) DESC

Show each gender code and the corresponding count of guests sorted by the count in descending order.

CREATE TABLE Guests (gender_code VARCHAR)

Mostrar cada código de gênero e a contagem correspondente de convidados classificados pela contagem em ordem decrescente.

SELECT COUNT(*) FROM Apartments WHERE NOT apt_id IN (SELECT apt_id FROM Apartment_Facilities)

How many apartments do not have any facility?

CREATE TABLE Apartment_Facilities (apt_id VARCHAR); CREATE TABLE Apartments (apt_id VARCHAR)

Quantos apartamentos não têm instalações?

SELECT T2.apt_number FROM Apartment_Bookings AS T1 JOIN Apartments AS T2 ON T1.apt_id = T2.apt_id WHERE T1.booking_status_code = "Confirmed" INTERSECT SELECT T2.apt_number FROM Apartment_Bookings AS T1 JOIN Apartments AS T2 ON T1.apt_id = T2.apt_id WHERE T1.booking_status_code = "Provisional"

Show the apartment numbers of apartments with bookings that have status code both "Provisional" and "Confirmed"

CREATE TABLE Apartments (apt_number VARCHAR, apt_id VARCHAR); CREATE TABLE Apartment_Bookings (apt_id VARCHAR, booking_status_code VARCHAR)

Mostrar os números de apartamentos de apartamentos com reservas que tenham código de status "Provisório" e "Confirmado"

SELECT T1.apt_number FROM Apartments AS T1 JOIN View_Unit_Status AS T2 ON T1.apt_id = T2.apt_id WHERE T2.available_yn = 0 INTERSECT SELECT T1.apt_number FROM Apartments AS T1 JOIN View_Unit_Status AS T2 ON T1.apt_id = T2.apt_id WHERE T2.available_yn = 1

Show the apartment numbers of apartments with unit status availability of both 0 and 1.

CREATE TABLE View_Unit_Status (apt_id VARCHAR, available_yn VARCHAR); CREATE TABLE Apartments (apt_number VARCHAR, apt_id VARCHAR)

Mostrar os números de apartamentos de apartamentos com disponibilidade de estado unitário de 0 e 1.

SELECT COUNT(*) FROM game WHERE season > 2007

How many games are held after season 2007?

CREATE TABLE game (season INTEGER)

Quantos jogos são realizados após a temporada de 2007?

SELECT Date FROM game ORDER BY home_team DESC

List the dates of games by the home team name in descending order.

CREATE TABLE game (Date VARCHAR, home_team VARCHAR)

Liste as datas dos jogos pelo nome da equipe da casa em ordem decrescente.

SELECT season, home_team, away_team FROM game

List the season, home team, away team of all the games.

CREATE TABLE game (season VARCHAR, home_team VARCHAR, away_team VARCHAR)

Listar a temporada, equipe da casa, equipe de distância de todos os jogos.

SELECT MAX(home_games), MIN(home_games), AVG(home_games) FROM stadium

What are the maximum, minimum and average home games each stadium held?

CREATE TABLE stadium (home_games INTEGER)

Quais são os jogos em casa máximos, mínimos e médios de cada estádio realizado?

SELECT average_attendance FROM stadium WHERE capacity_percentage > 100

What is the average attendance of stadiums with capacity percentage higher than 100%?

CREATE TABLE stadium (average_attendance VARCHAR, capacity_percentage INTEGER)

Qual é a frequência média de estádios com percentual de capacidade superior a 100%?

SELECT player, number_of_matches, SOURCE FROM injury_accident WHERE injury <> 'Knee problem'

What are the player name, number of matches, and information source for players who do not suffer from injury of 'Knee problem'?

CREATE TABLE injury_accident (player VARCHAR, number_of_matches VARCHAR, SOURCE VARCHAR, injury VARCHAR)

Qual é o nome do jogador, o número de partidas e a fonte de informação para os jogadores que não sofrem de lesão do 'problema do joelho'?

SELECT T1.season FROM game AS T1 JOIN injury_accident AS T2 ON T1.id = T2.game_id WHERE T2.player = 'Walter Samuel'

What is the season of the game which causes the player 'Walter Samuel' to get injured?

CREATE TABLE injury_accident (game_id VARCHAR, player VARCHAR); CREATE TABLE game (season VARCHAR, id VARCHAR)

Qual é a temporada do jogo que faz com que o jogador 'Walter Samuel' se machuque?

SELECT T1.id, T1.score, T1.date FROM game AS T1 JOIN injury_accident AS T2 ON T2.game_id = T1.id GROUP BY T1.id HAVING COUNT(*) >= 2

What are the ids, scores, and dates of the games which caused at least two injury accidents?

CREATE TABLE game (id VARCHAR, score VARCHAR, date VARCHAR); CREATE TABLE injury_accident (game_id VARCHAR)

Quais são as IDs, pontuações e datas dos jogos que causaram pelo menos dois acidentes de lesão?

SELECT T1.id, T1.name FROM stadium AS T1 JOIN game AS T2 ON T1.id = T2.stadium_id JOIN injury_accident AS T3 ON T2.id = T3.game_id GROUP BY T1.id ORDER BY COUNT(*) DESC LIMIT 1

What are the id and name of the stadium where the most injury accidents happened?

CREATE TABLE stadium (id VARCHAR, name VARCHAR); CREATE TABLE injury_accident (game_id VARCHAR); CREATE TABLE game (stadium_id VARCHAR, id VARCHAR)

Qual é o ID e o nome do estádio onde ocorreram os acidentes mais lesionados?

SELECT T1.season, T2.name FROM game AS T1 JOIN stadium AS T2 ON T1.stadium_id = T2.id JOIN injury_accident AS T3 ON T1.id = T3.game_id WHERE T3.injury = 'Foot injury' OR T3.injury = 'Knee problem'

In which season and which stadium did any player have an injury of 'Foot injury' or 'Knee problem'?

CREATE TABLE stadium (name VARCHAR, id VARCHAR); CREATE TABLE injury_accident (game_id VARCHAR, injury VARCHAR); CREATE TABLE game (season VARCHAR, stadium_id VARCHAR, id VARCHAR)

Em qual temporada e em qual estádio qualquer jogador teve uma lesão de "ferida no pé" ou "problema do joelho"?

SELECT COUNT(DISTINCT SOURCE) FROM injury_accident

How many different kinds of information sources are there for injury accidents?

CREATE TABLE injury_accident (SOURCE VARCHAR)

Quantos tipos diferentes de fontes de informação existem para acidentes de lesão?

SELECT COUNT(*) FROM game WHERE NOT id IN (SELECT game_id FROM injury_accident)

How many games are free of injury accidents?

CREATE TABLE injury_accident (id VARCHAR, game_id VARCHAR); CREATE TABLE game (id VARCHAR, game_id VARCHAR)

Quantos jogos estão livres de acidentes de lesão?

SELECT COUNT(DISTINCT T1.injury) FROM injury_accident AS T1 JOIN game AS T2 ON T1.game_id = T2.id WHERE T2.season > 2010

How many distinct kinds of injuries happened after season 2010?

CREATE TABLE injury_accident (injury VARCHAR, game_id VARCHAR); CREATE TABLE game (id VARCHAR, season INTEGER)

Quantos tipos distintos de lesões aconteceram após a temporada de 2010?

SELECT T2.name FROM game AS T1 JOIN stadium AS T2 ON T1.stadium_id = T2.id JOIN injury_accident AS T3 ON T1.id = T3.game_id WHERE T3.player = 'Walter Samuel' INTERSECT SELECT T2.name FROM game AS T1 JOIN stadium AS T2 ON T1.stadium_id = T2.id JOIN injury_accident AS T3 ON T1.id = T3.game_id WHERE T3.player = 'Thiago Motta'

List the name of the stadium where both the player 'Walter Samuel' and the player 'Thiago Motta' got injured.

CREATE TABLE stadium (name VARCHAR, id VARCHAR); CREATE TABLE game (stadium_id VARCHAR, id VARCHAR); CREATE TABLE injury_accident (game_id VARCHAR, player VARCHAR)

Liste o nome do estádio onde o jogador 'Walter Samuel' e o jogador 'Thiago Motta' ficaram feridos.

SELECT name, average_attendance, total_attendance FROM stadium EXCEPT SELECT T2.name, T2.average_attendance, T2.total_attendance FROM game AS T1 JOIN stadium AS T2 ON T1.stadium_id = T2.id JOIN injury_accident AS T3 ON T1.id = T3.game_id

Show the name, average attendance, total attendance for stadiums where no accidents happened.

CREATE TABLE stadium (name VARCHAR, average_attendance VARCHAR, total_attendance VARCHAR, id VARCHAR); CREATE TABLE stadium (name VARCHAR, average_attendance VARCHAR, total_attendance VARCHAR); CREATE TABLE game (stadium_id VARCHAR, id VARCHAR); CREATE TABLE injury_accident (game_id VARCHAR)

Mostre o nome, a média de atendimento, a assistência total para estádios onde nenhum acidente aconteceu.

SELECT name FROM stadium WHERE name LIKE "%Bank%"

Which stadium name contains the substring "Bank"?

CREATE TABLE stadium (name VARCHAR)

Qual nome do estádio contém a substring "Banco"?

SELECT T1.id, COUNT(*) FROM stadium AS T1 JOIN game AS T2 ON T1.id = T2.stadium_id GROUP BY T1.id

How many games has each stadium held?

CREATE TABLE stadium (id VARCHAR); CREATE TABLE game (stadium_id VARCHAR)

Quantos jogos cada estádio tem realizado?

SELECT T1.date, T2.player FROM game AS T1 JOIN injury_accident AS T2 ON T1.id = T2.game_id ORDER BY T1.season DESC

For each injury accident, find the date of the game and the name of the injured player in the game, and sort the results in descending order of game season.

CREATE TABLE game (date VARCHAR, id VARCHAR, season VARCHAR); CREATE TABLE injury_accident (player VARCHAR, game_id VARCHAR)

Para cada acidente de lesão, encontrar a data do jogo e o nome do jogador ferido no jogo, e classificar os resultados em ordem decrescente da temporada de jogo.

SELECT T1.name, T2.name FROM Country AS T1 JOIN League AS T2 ON T1.id = T2.country_id

List all country and league names.

CREATE TABLE League (name VARCHAR, country_id VARCHAR); CREATE TABLE Country (name VARCHAR, id VARCHAR)

Listar todos os nomes de países e ligas.

SELECT COUNT(*) FROM Country AS T1 JOIN League AS T2 ON T1.id = T2.country_id WHERE T1.name = "England"

How many leagues are there in England?

CREATE TABLE League (country_id VARCHAR); CREATE TABLE Country (id VARCHAR, name VARCHAR)

Quantas ligas existem na Inglaterra?

SELECT AVG(weight) FROM Player

What is the average weight of all players?

CREATE TABLE Player (weight INTEGER)

Qual é o peso médio de todos os jogadores?

SELECT MAX(weight), MIN(weight) FROM Player

What is the maximum and minimum height of all players?

CREATE TABLE Player (weight INTEGER)

Qual é a altura máxima e mínima de todos os jogadores?

SELECT DISTINCT T1.player_name FROM Player AS T1 JOIN Player_Attributes AS T2 ON T1.player_api_id = T2.player_api_id WHERE T2.overall_rating > (SELECT AVG(overall_rating) FROM Player_Attributes)

List all player names who have an overall rating higher than the average.

CREATE TABLE Player (player_name VARCHAR, player_api_id VARCHAR); CREATE TABLE Player_Attributes (overall_rating INTEGER); CREATE TABLE Player_Attributes (player_api_id VARCHAR, overall_rating INTEGER)

Listar todos os nomes de jogadores que têm uma classificação geral superior à média.

SELECT DISTINCT T1.player_name FROM Player AS T1 JOIN Player_Attributes AS T2 ON T1.player_api_id = T2.player_api_id WHERE T2.dribbling = (SELECT MAX(overall_rating) FROM Player_Attributes)

What are the names of players who have the best dribbling?

CREATE TABLE Player (player_name VARCHAR, player_api_id VARCHAR); CREATE TABLE Player_Attributes (player_api_id VARCHAR, dribbling VARCHAR); CREATE TABLE Player_Attributes (overall_rating INTEGER)

Quais são os nomes dos jogadores que têm os melhores dribles?

SELECT DISTINCT T1.player_name FROM Player AS T1 JOIN Player_Attributes AS T2 ON T1.player_api_id = T2.player_api_id WHERE T2.crossing > 90 AND T2.preferred_foot = "right"

List the names of all players who have a crossing score higher than 90 and prefer their right foot.

CREATE TABLE Player (player_name VARCHAR, player_api_id VARCHAR); CREATE TABLE Player_Attributes (player_api_id VARCHAR, crossing VARCHAR, preferred_foot VARCHAR)

Listar os nomes de todos os jogadores que têm uma pontuação de cruzamento superior a 90 e preferem o seu pé direito.

SELECT DISTINCT T1.player_name FROM Player AS T1 JOIN Player_Attributes AS T2 ON T1.player_api_id = T2.player_api_id WHERE T2.preferred_foot = "left" AND T2.overall_rating >= 85 AND T2.overall_rating <= 90

List the names of all left-footed players who have overall rating between 85 and 90.

CREATE TABLE Player (player_name VARCHAR, player_api_id VARCHAR); CREATE TABLE Player_Attributes (player_api_id VARCHAR, overall_rating VARCHAR, preferred_foot VARCHAR)

Listar os nomes de todos os jogadores de pé esquerdo que têm classificação geral entre 85 e 90.

SELECT preferred_foot, AVG(overall_rating) FROM Player_Attributes GROUP BY preferred_foot

What is the average rating for right-footed players and left-footed players?

CREATE TABLE Player_Attributes (preferred_foot VARCHAR, overall_rating INTEGER)

Qual é a classificação média para jogadores com o pé direito e jogadores com o pé esquerdo?

SELECT preferred_foot, COUNT(*) FROM Player_Attributes WHERE overall_rating > 80 GROUP BY preferred_foot

Of all players with an overall rating greater than 80, how many are right-footed and left-footed?

CREATE TABLE Player_Attributes (preferred_foot VARCHAR, overall_rating INTEGER)

De todos os jogadores com uma classificação geral superior a 80, quantos são destros e destros?

SELECT player_api_id FROM Player WHERE height >= 180 INTERSECT SELECT player_api_id FROM Player_Attributes WHERE overall_rating > 85

List all of the player ids with a height of at least 180cm and an overall rating higher than 85.

CREATE TABLE Player_Attributes (player_api_id VARCHAR, height VARCHAR, overall_rating INTEGER); CREATE TABLE Player (player_api_id VARCHAR, height VARCHAR, overall_rating INTEGER)

Liste todos os IDs do jogador com uma altura de pelo menos 180 cm e uma classificação geral superior a 85.

SELECT player_api_id FROM Player WHERE height >= 180 AND height <= 190 INTERSECT SELECT player_api_id FROM Player_Attributes WHERE preferred_foot = "left"

List all of the ids for left-footed players with a height between 180cm and 190cm.

CREATE TABLE Player_Attributes (player_api_id VARCHAR, preferred_foot VARCHAR, height VARCHAR); CREATE TABLE Player (player_api_id VARCHAR, preferred_foot VARCHAR, height VARCHAR)

Liste todos os IDs para jogadores de pé esquerdo com uma altura entre 180cm e 190cm.

SELECT DISTINCT T1.player_name FROM Player AS T1 JOIN Player_Attributes AS T2 ON T1.player_api_id = T2.player_api_id ORDER BY overall_rating DESC LIMIT 3

Who are the top 3 players in terms of overall rating?

CREATE TABLE Player (player_name VARCHAR, player_api_id VARCHAR); CREATE TABLE Player_Attributes (player_api_id VARCHAR)

Quem são os 3 melhores jogadores em termos de classificação geral?

SELECT DISTINCT T1.player_name, T1.birthday FROM Player AS T1 JOIN Player_Attributes AS T2 ON T1.player_api_id = T2.player_api_id ORDER BY potential DESC LIMIT 5

List the names and birthdays of the top five players in terms of potential.

CREATE TABLE Player_Attributes (player_api_id VARCHAR); CREATE TABLE Player (player_name VARCHAR, birthday VARCHAR, player_api_id VARCHAR)

Liste os nomes e aniversários dos cinco melhores jogadores em termos de potencial.

SELECT COUNT(*) FROM performance

How many performances are there?

CREATE TABLE performance (Id VARCHAR)

Quantas performances existem?

SELECT HOST FROM performance ORDER BY Attendance

List the hosts of performances in ascending order of attendance.

CREATE TABLE performance (HOST VARCHAR, Attendance VARCHAR)

Listar as hostes de performances em ordem crescente de atendimento.

SELECT Date, LOCATION FROM performance

What are the dates and locations of performances?

CREATE TABLE performance (Date VARCHAR, LOCATION VARCHAR)

Quais são as datas e locais das apresentações?

SELECT Attendance FROM performance WHERE LOCATION = "TD Garden" OR LOCATION = "Bell Centre"

Show the attendances of the performances at location "TD Garden" or "Bell Centre"

CREATE TABLE performance (Attendance VARCHAR, LOCATION VARCHAR)

Mostre os atendimentos das apresentações no local "TD Garden" ou "Bell Centre"

SELECT AVG(Attendance) FROM performance

What is the average number of attendees for performances?

CREATE TABLE performance (Attendance INTEGER)

Qual é o número médio de participantes para performances?

SELECT Date FROM performance ORDER BY Attendance DESC LIMIT 1

What is the date of the performance with the highest number of attendees?

CREATE TABLE performance (Date VARCHAR, Attendance VARCHAR)

Qual é a data do desempenho com o maior número de participantes?

SELECT LOCATION, COUNT(*) FROM performance GROUP BY LOCATION

Show different locations and the number of performances at each location.

CREATE TABLE performance (LOCATION VARCHAR)

Mostre diferentes locais e o número de apresentações em cada local.

SELECT LOCATION FROM performance GROUP BY LOCATION ORDER BY COUNT(*) DESC LIMIT 1

Show the most common location of performances.

CREATE TABLE performance (LOCATION VARCHAR)

Mostre a localização mais comum das performances.

SELECT LOCATION FROM performance GROUP BY LOCATION HAVING COUNT(*) >= 2

Show the locations that have at least two performances.

CREATE TABLE performance (LOCATION VARCHAR)

Mostre os locais que têm pelo menos duas apresentações.

SELECT LOCATION FROM performance WHERE Attendance > 2000 INTERSECT SELECT LOCATION FROM performance WHERE Attendance < 1000

Show the locations that have both performances with more than 2000 attendees and performances with less than 1000 attendees.

CREATE TABLE performance (LOCATION VARCHAR, Attendance INTEGER)

Mostre os locais que têm ambas as apresentações com mais de 2000 participantes e performances com menos de 1000 participantes.

SELECT T2.Name, T3.Location FROM member_attendance AS T1 JOIN member AS T2 ON T1.Member_ID = T2.Member_ID JOIN performance AS T3 ON T1.Performance_ID = T3.Performance_ID

Show the names of members and the location of the performances they attended.

CREATE TABLE performance (Location VARCHAR, Performance_ID VARCHAR); CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR); CREATE TABLE member_attendance (Member_ID VARCHAR, Performance_ID VARCHAR)

Mostre os nomes dos membros e a localização das apresentações que eles assistiram.

SELECT T2.Name, T3.Location FROM member_attendance AS T1 JOIN member AS T2 ON T1.Member_ID = T2.Member_ID JOIN performance AS T3 ON T1.Performance_ID = T3.Performance_ID ORDER BY T2.Name

Show the names of members and the location of performances they attended in ascending alphabetical order of their names.

CREATE TABLE performance (Location VARCHAR, Performance_ID VARCHAR); CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR); CREATE TABLE member_attendance (Member_ID VARCHAR, Performance_ID VARCHAR)

Mostre os nomes dos membros e a localização das performances que eles assistiram em ordem alfabética ascendente de seus nomes.

SELECT T3.Date FROM member_attendance AS T1 JOIN member AS T2 ON T1.Member_ID = T2.Member_ID JOIN performance AS T3 ON T1.Performance_ID = T3.Performance_ID WHERE T2.Role = "Violin"

Show the dates of performances with attending members whose roles are "Violin".

CREATE TABLE performance (Date VARCHAR, Performance_ID VARCHAR); CREATE TABLE member (Member_ID VARCHAR, Role VARCHAR); CREATE TABLE member_attendance (Member_ID VARCHAR, Performance_ID VARCHAR)

Mostre as datas das apresentações com os membros presentes cujos papéis são "Violino".

SELECT T2.Name, T3.Date FROM member_attendance AS T1 JOIN member AS T2 ON T1.Member_ID = T2.Member_ID JOIN performance AS T3 ON T1.Performance_ID = T3.Performance_ID ORDER BY T3.Attendance DESC

Show the names of members and the dates of performances they attended in descending order of attendance of the performances.

CREATE TABLE performance (Date VARCHAR, Performance_ID VARCHAR, Attendance VARCHAR); CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR); CREATE TABLE member_attendance (Member_ID VARCHAR, Performance_ID VARCHAR)

Mostre os nomes dos membros e as datas das apresentações que eles assistiram em ordem decrescente de atendimento das apresentações.

SELECT Name FROM member WHERE NOT Member_ID IN (SELECT Member_ID FROM member_attendance)

List the names of members who did not attend any performance.

CREATE TABLE member (Name VARCHAR, Member_ID VARCHAR); CREATE TABLE member_attendance (Name VARCHAR, Member_ID VARCHAR)

Listar os nomes dos membros que não compareceram a nenhuma performance.

SELECT DISTINCT building FROM classroom WHERE capacity > 50

Find the buildings which have rooms with capacity more than 50.

CREATE TABLE classroom (building VARCHAR, capacity INTEGER)

Encontre os edifícios que têm quartos com capacidade superior a 50.

SELECT COUNT(*) FROM classroom WHERE building <> 'Lamberton'

Count the number of rooms that are not in the Lamberton building.

CREATE TABLE classroom (building VARCHAR)

Conte o número de quartos que não estão no edifício Lamberton.

SELECT dept_name, building FROM department WHERE budget > (SELECT AVG(budget) FROM department)

What is the name and building of the departments whose budget is more than the average budget?

CREATE TABLE department (dept_name VARCHAR, building VARCHAR, budget INTEGER)

Qual é o nome e a construção dos departamentos cujo orçamento é mais do que o orçamento médio?

SELECT building, room_number FROM classroom WHERE capacity BETWEEN 50 AND 100

Find the room number of the rooms which can sit 50 to 100 students and their buildings.

CREATE TABLE classroom (building VARCHAR, room_number VARCHAR, capacity INTEGER)

Encontre o número do quarto dos quartos que podem acomodar de 50 a 100 alunos e seus edifícios.

SELECT dept_name, building FROM department ORDER BY budget DESC LIMIT 1

Find the name and building of the department with the highest budget.

CREATE TABLE department (dept_name VARCHAR, building VARCHAR, budget VARCHAR)

Encontre o nome e a construção do departamento com o orçamento mais alto.

SELECT name FROM student WHERE dept_name = 'History' ORDER BY tot_cred DESC LIMIT 1

What is the name of the student who has the highest total credits in the History department.

CREATE TABLE student (name VARCHAR, dept_name VARCHAR, tot_cred VARCHAR)

Qual é o nome do aluno que tem o maior total de créditos no departamento de História?

SELECT COUNT(*) FROM classroom WHERE building = 'Lamberton'

How many rooms does the Lamberton building have?

CREATE TABLE classroom (building VARCHAR)

Quantos quartos tem o edifício Lamberton?

SELECT COUNT(DISTINCT s_id) FROM advisor

How many students have advisors?

CREATE TABLE advisor (s_id VARCHAR)

Quantos alunos têm orientadores?

SELECT COUNT(DISTINCT dept_name) FROM course

How many departments offer courses?

CREATE TABLE course (dept_name VARCHAR)

Quantos departamentos oferecem cursos?

SELECT COUNT(DISTINCT course_id) FROM course WHERE dept_name = 'Physics'

How many different courses offered by Physics department?

CREATE TABLE course (course_id VARCHAR, dept_name VARCHAR)

Quantos cursos diferentes oferecidos pelo departamento de Física?

SELECT T1.title FROM course AS T1 JOIN prereq AS T2 ON T1.course_id = T2.course_id GROUP BY T2.course_id HAVING COUNT(*) = 2

Find the title of courses that have two prerequisites?

CREATE TABLE prereq (course_id VARCHAR); CREATE TABLE course (title VARCHAR, course_id VARCHAR)

Encontrar o título dos cursos que têm dois pré-requisitos?

SELECT T1.title, T1.credits, T1.dept_name FROM course AS T1 JOIN prereq AS T2 ON T1.course_id = T2.course_id GROUP BY T2.course_id HAVING COUNT(*) > 1

Find the title, credit, and department name of courses that have more than one prerequisites?

CREATE TABLE prereq (course_id VARCHAR); CREATE TABLE course (title VARCHAR, credits VARCHAR, dept_name VARCHAR, course_id VARCHAR)

Encontre o título, o crédito e o nome do departamento dos cursos que têm mais de um pré-requisitos?

SELECT COUNT(*) FROM course WHERE NOT course_id IN (SELECT course_id FROM prereq)

How many courses that do not have prerequisite?

CREATE TABLE prereq (course_id VARCHAR); CREATE TABLE course (course_id VARCHAR)

Quantos cursos não têm pré-requisito?

SELECT title FROM course WHERE NOT course_id IN (SELECT course_id FROM prereq)

Find the name of the courses that do not have any prerequisite?

CREATE TABLE prereq (title VARCHAR, course_id VARCHAR); CREATE TABLE course (title VARCHAR, course_id VARCHAR)

Encontrar o nome dos cursos que não têm qualquer pré-requisito?

SELECT COUNT(DISTINCT id) FROM teaches

How many different instructors have taught some course?

CREATE TABLE teaches (id VARCHAR)

Quantos instrutores diferentes ensinaram algum curso?

SELECT SUM(budget) FROM department WHERE dept_name = 'Marketing' OR dept_name = 'Finance'

Find the total budgets of the Marketing or Finance department.

CREATE TABLE department (budget INTEGER, dept_name VARCHAR)

Encontre os orçamentos totais do departamento de Marketing ou Finanças.

SELECT dept_name FROM instructor WHERE name LIKE '%Soisalon%'

Find the department name of the instructor whose name contains 'Soisalon'.

CREATE TABLE instructor (dept_name VARCHAR, name VARCHAR)

Encontre o nome do departamento do instrutor cujo nome contém 'Soisalon'.

SELECT COUNT(*) FROM classroom WHERE building = 'Lamberton' AND capacity < 50

How many rooms whose capacity is less than 50 does the Lamberton building have?

CREATE TABLE classroom (building VARCHAR, capacity VARCHAR)

Quantos quartos com capacidade inferior a 50 tem o edifício Lamberton?

SELECT dept_name, budget FROM department WHERE budget > (SELECT AVG(budget) FROM department)

Find the name and budget of departments whose budgets are more than the average budget.

CREATE TABLE department (dept_name VARCHAR, budget INTEGER)

Encontre o nome e o orçamento dos departamentos cujos orçamentos são mais do que o orçamento médio.

SELECT name FROM instructor WHERE dept_name = 'Statistics' ORDER BY salary LIMIT 1

what is the name of the instructor who is in Statistics department and earns the lowest salary?

CREATE TABLE instructor (name VARCHAR, dept_name VARCHAR, salary VARCHAR)

Qual é o nome do instrutor que está no departamento de Estatística e ganha o salário mais baixo?

SELECT title FROM course WHERE dept_name = 'Statistics' INTERSECT SELECT title FROM course WHERE dept_name = 'Psychology'

Find the title of course that is provided by both Statistics and Psychology departments.

CREATE TABLE course (title VARCHAR, dept_name VARCHAR)

Encontre o título do curso que é fornecido por ambos os departamentos de Estatística e Psicologia.

SELECT title FROM course WHERE dept_name = 'Statistics' EXCEPT SELECT title FROM course WHERE dept_name = 'Psychology'

Find the title of course that is provided by Statistics but not Psychology departments.

CREATE TABLE course (title VARCHAR, dept_name VARCHAR)

Encontre o título do curso que é fornecido pelos departamentos de Estatística, mas não de Psicologia.

SELECT id FROM teaches WHERE semester = 'Fall' AND YEAR = 2009 EXCEPT SELECT id FROM teaches WHERE semester = 'Spring' AND YEAR = 2010

Find the id of instructors who taught a class in Fall 2009 but not in Spring 2010.

CREATE TABLE teaches (id VARCHAR, semester VARCHAR, YEAR VARCHAR)

Encontre o id de instrutores que ensinaram uma aula no outono de 2009, mas não na primavera de 2010.

SELECT DISTINCT T1.name FROM student AS T1 JOIN takes AS T2 ON T1.id = T2.id WHERE YEAR = 2009 OR YEAR = 2010

Find the name of students who took any class in the years of 2009 and 2010.

CREATE TABLE student (name VARCHAR, id VARCHAR); CREATE TABLE takes (id VARCHAR)

Encontre o nome dos alunos que fizeram qualquer aula nos anos de 2009 e 2010.

SELECT dept_name FROM course GROUP BY dept_name ORDER BY COUNT(*) DESC LIMIT 3

Find the names of the top 3 departments that provide the largest amount of courses?

CREATE TABLE course (dept_name VARCHAR)

Encontre os nomes dos 3 principais departamentos que oferecem a maior quantidade de cursos?

SELECT dept_name FROM course GROUP BY dept_name ORDER BY SUM(credits) DESC LIMIT 1

Find the name of the department that offers the highest total credits?

CREATE TABLE course (dept_name VARCHAR, credits INTEGER)

Encontre o nome do departamento que oferece o maior total de créditos?

SELECT title FROM course ORDER BY title, credits

List the names of all courses ordered by their titles and credits.

CREATE TABLE course (title VARCHAR, credits VARCHAR)

Listar os nomes de todos os cursos ordenados por seus títulos e créditos.

SELECT dept_name FROM department ORDER BY budget LIMIT 1

Which department has the lowest budget?

CREATE TABLE department (dept_name VARCHAR, budget VARCHAR)

Qual departamento tem o orçamento mais baixo?

SELECT dept_name, building FROM department ORDER BY budget DESC

List the names and buildings of all departments sorted by the budget from large to small.

CREATE TABLE department (dept_name VARCHAR, building VARCHAR, budget VARCHAR)

Liste os nomes e edifícios de todos os departamentos classificados pelo orçamento de grande para pequeno.

SELECT name FROM instructor ORDER BY salary DESC LIMIT 1

Who is the instructor with the highest salary?

CREATE TABLE instructor (name VARCHAR, salary VARCHAR)

Quem é o instrutor com o salário mais alto?

SELECT * FROM instructor ORDER BY salary

List the information of all instructors ordered by their salary in ascending order.

CREATE TABLE instructor (salary VARCHAR)

Liste as informações de todos os instrutores ordenados por seu salário em ordem crescente.

SELECT name, dept_name FROM student ORDER BY tot_cred

Find the name of the students and their department names sorted by their total credits in ascending order.

CREATE TABLE student (name VARCHAR, dept_name VARCHAR, tot_cred VARCHAR)

Encontre o nome dos alunos e seus nomes de departamento classificados por seus créditos totais em ordem crescente.

SELECT T1.title, T3.name FROM course AS T1 JOIN teaches AS T2 ON T1.course_id = T2.course_id JOIN instructor AS T3 ON T2.id = T3.id WHERE YEAR = 2008 ORDER BY T1.title

list in alphabetic order all course names and their instructors' names in year 2008.

CREATE TABLE course (title VARCHAR, course_id VARCHAR); CREATE TABLE teaches (course_id VARCHAR, id VARCHAR); CREATE TABLE instructor (name VARCHAR, id VARCHAR)

lista em ordem alfabética todos os nomes dos cursos e nomes de seus instrutores no ano de 2008.