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SELECT card_type_code FROM Customers_cards GROUP BY card_type_code ORDER BY COUNT(*) DESC LIMIT 1

What is the card type code with most number of cards?

CREATE TABLE Customers_cards (card_type_code VARCHAR)

Qual é o código do tipo de cartão com o maior número de cartões?

SELECT card_type_code FROM Customers_cards GROUP BY card_type_code HAVING COUNT(*) >= 5

Show card type codes with at least 5 cards.

CREATE TABLE Customers_cards (card_type_code VARCHAR)

Mostrar códigos de tipo de cartão com pelo menos 5 cartões.

SELECT card_type_code, COUNT(DISTINCT customer_id) FROM Customers_cards GROUP BY card_type_code

Show all card type codes and the number of customers holding cards in each type.

CREATE TABLE Customers_cards (card_type_code VARCHAR, customer_id VARCHAR)

Mostre todos os códigos de tipo de cartão e o número de clientes que possuem cartões em cada tipo.

SELECT customer_id, customer_first_name FROM Customers EXCEPT SELECT T1.customer_id, T2.customer_first_name FROM Customers_cards AS T1 JOIN Customers AS T2 ON T1.customer_id = T2.customer_id WHERE card_type_code = "Credit"

Show the customer ids and firstname without a credit card.

CREATE TABLE Customers_cards (customer_id VARCHAR); CREATE TABLE Customers (customer_first_name VARCHAR, customer_id VARCHAR); CREATE TABLE Customers (customer_id VARCHAR, customer_first_name VARCHAR, card_type_code VARCHAR)

Mostre as IDs do cliente e o primeiro nome sem um cartão de crédito.

SELECT DISTINCT card_type_code FROM Customers_Cards

Show all card type codes.

CREATE TABLE Customers_Cards (card_type_code VARCHAR)

Mostrar todos os códigos de tipo de cartão.

SELECT COUNT(DISTINCT card_type_code) FROM Customers_Cards

Show the number of card types.

CREATE TABLE Customers_Cards (card_type_code VARCHAR)

Mostrar o número de tipos de cartão.

SELECT DISTINCT transaction_type FROM Financial_Transactions

Show all transaction types.

CREATE TABLE Financial_Transactions (transaction_type VARCHAR)

Mostrar todos os tipos de transação.

SELECT COUNT(DISTINCT transaction_type) FROM Financial_Transactions

Show the number of transaction types.

CREATE TABLE Financial_Transactions (transaction_type VARCHAR)

Mostrar o número de tipos de transação.

SELECT AVG(transaction_amount), SUM(transaction_amount) FROM Financial_transactions

What is the average and total transaction amount?

CREATE TABLE Financial_transactions (transaction_amount INTEGER)

Qual é o valor médio e total da transação?

SELECT T2.card_type_code, COUNT(*) FROM Financial_transactions AS T1 JOIN Customers_cards AS T2 ON T1.card_id = T2.card_id GROUP BY T2.card_type_code

Show the card type codes and the number of transactions.

CREATE TABLE Financial_transactions (card_id VARCHAR); CREATE TABLE Customers_cards (card_type_code VARCHAR, card_id VARCHAR)

Mostre os códigos de tipo de cartão e o número de transações.

SELECT transaction_type, COUNT(*) FROM Financial_transactions GROUP BY transaction_type

Show the transaction type and the number of transactions.

CREATE TABLE Financial_transactions (transaction_type VARCHAR)

Mostrar o tipo de transação e o número de transações.

SELECT transaction_type FROM Financial_transactions GROUP BY transaction_type ORDER BY SUM(transaction_amount) DESC LIMIT 1

What is the transaction type that has processed the greatest total amount in transactions?

CREATE TABLE Financial_transactions (transaction_type VARCHAR, transaction_amount INTEGER)

Qual é o tipo de transação que processou o maior valor total em transações?

SELECT account_id, COUNT(*) FROM Financial_transactions GROUP BY account_id

Show the account id and the number of transactions for each account

CREATE TABLE Financial_transactions (account_id VARCHAR)

Mostrar o ID da conta e o número de transações para cada conta

SELECT COUNT(*) FROM track

How many tracks do we have?

CREATE TABLE track (Id VARCHAR)

Quantas faixas temos?

SELECT name, LOCATION FROM track

Show the name and location for all tracks.

CREATE TABLE track (name VARCHAR, LOCATION VARCHAR)

Mostre o nome e a localização de todas as faixas.

SELECT name, seating FROM track WHERE year_opened > 2000 ORDER BY seating

Show names and seatings, ordered by seating for all tracks opened after 2000.

CREATE TABLE track (name VARCHAR, seating VARCHAR, year_opened INTEGER)

Mostrar nomes e assentos, ordenados por assentos para todas as faixas abertas após 2000.

SELECT name, LOCATION, seating FROM track ORDER BY year_opened DESC LIMIT 1

What is the name, location and seating for the most recently opened track?

CREATE TABLE track (name VARCHAR, LOCATION VARCHAR, seating VARCHAR, year_opened VARCHAR)

Qual é o nome, localização e assentos para a pista mais recentemente aberta?

SELECT MIN(seating), MAX(seating), AVG(seating) FROM track

What is the minimum, maximum, and average seating for all tracks.

CREATE TABLE track (seating INTEGER)

Qual é o assento mínimo, máximo e médio para todas as faixas.

SELECT name, LOCATION, year_opened FROM track WHERE seating > (SELECT AVG(seating) FROM track)

Show the name, location, open year for all tracks with a seating higher than the average.

CREATE TABLE track (name VARCHAR, LOCATION VARCHAR, year_opened VARCHAR, seating INTEGER)

Mostre o nome, a localização, o ano aberto para todas as faixas com um assento superior à média.

SELECT DISTINCT LOCATION FROM track

What are distinct locations where tracks are located?

CREATE TABLE track (LOCATION VARCHAR)

Quais são os locais onde as pistas estão localizadas?

SELECT COUNT(*) FROM race

How many races are there?

CREATE TABLE race (Id VARCHAR)

Quantas raças existem?

SELECT DISTINCT CLASS FROM race

What are the distinct classes that races can have?

CREATE TABLE race (CLASS VARCHAR)

Quais são as classes distintas que as raças podem ter?

SELECT name, CLASS, date FROM race

Show name, class, and date for all races.

CREATE TABLE race (name VARCHAR, CLASS VARCHAR, date VARCHAR)

Mostrar nome, classe e data para todas as raças.

SELECT CLASS, COUNT(*) FROM race GROUP BY CLASS

Show the race class and number of races in each class.

CREATE TABLE race (CLASS VARCHAR)

Mostre a classe de corrida e o número de corridas em cada classe.

SELECT CLASS FROM race GROUP BY CLASS ORDER BY COUNT(*) DESC LIMIT 1

What is the race class with most number of races.

CREATE TABLE race (CLASS VARCHAR)

Qual é a classe de corrida com o maior número de raças?

SELECT CLASS FROM race GROUP BY CLASS HAVING COUNT(*) >= 2

List the race class with at least two races.

CREATE TABLE race (CLASS VARCHAR)

Liste a classe de corrida com pelo menos duas corridas.

SELECT name FROM track EXCEPT SELECT T2.name FROM race AS T1 JOIN track AS T2 ON T1.track_id = T2.track_id WHERE T1.class = 'GT'

What are the names for tracks without a race in class 'GT'.

CREATE TABLE race (track_id VARCHAR, class VARCHAR); CREATE TABLE track (name VARCHAR); CREATE TABLE track (name VARCHAR, track_id VARCHAR)

Quais são os nomes das pistas sem uma corrida na classe 'GT'.

SELECT name FROM track WHERE NOT track_id IN (SELECT track_id FROM race)

Show all track names that have had no races.

CREATE TABLE track (name VARCHAR, track_id VARCHAR); CREATE TABLE race (name VARCHAR, track_id VARCHAR)

Mostre todos os nomes de pistas que não tiveram corridas.

SELECT year_opened FROM track WHERE seating BETWEEN 4000 AND 5000

Show year where a track with a seating at least 5000 opened and a track with seating no more than 4000 opened.

CREATE TABLE track (year_opened VARCHAR, seating INTEGER)

Show ano onde uma pista com um assento de pelo menos 5000 aberto e uma pista com assentos não mais de 4000 aberto.

SELECT T2.name, COUNT(*) FROM race AS T1 JOIN track AS T2 ON T1.track_id = T2.track_id GROUP BY T1.track_id

Show the name of track and the number of races in each track.

CREATE TABLE track (name VARCHAR, track_id VARCHAR); CREATE TABLE race (track_id VARCHAR)

Mostre o nome da pista e o número de corridas em cada pista.

SELECT T2.name FROM race AS T1 JOIN track AS T2 ON T1.track_id = T2.track_id GROUP BY T1.track_id ORDER BY COUNT(*) DESC LIMIT 1

Show the name of track with most number of races.

CREATE TABLE track (name VARCHAR, track_id VARCHAR); CREATE TABLE race (track_id VARCHAR)

Mostre o nome da pista com o maior número de corridas.

SELECT T1.name, T1.date, T2.name FROM race AS T1 JOIN track AS T2 ON T1.track_id = T2.track_id

Show the name and date for each race and its track name.

CREATE TABLE race (name VARCHAR, date VARCHAR, track_id VARCHAR); CREATE TABLE track (name VARCHAR, track_id VARCHAR)

Mostre o nome e a data de cada corrida e seu nome de pista.

SELECT T2.name, T2.location FROM race AS T1 JOIN track AS T2 ON T1.track_id = T2.track_id GROUP BY T1.track_id HAVING COUNT(*) = 1

Show the name and location of track with 1 race.

CREATE TABLE race (track_id VARCHAR); CREATE TABLE track (name VARCHAR, location VARCHAR, track_id VARCHAR)

Mostre o nome e a localização da pista com 1 corrida.

SELECT LOCATION FROM track WHERE seating > 90000 INTERSECT SELECT LOCATION FROM track WHERE seating < 70000

Find the locations where have both tracks with more than 90000 seats and tracks with less than 70000 seats.

CREATE TABLE track (LOCATION VARCHAR, seating INTEGER)

Encontre os locais onde têm ambas as faixas com mais de 90000 assentos e faixas com menos de 70000 assentos.

SELECT COUNT(*) FROM member WHERE Membership_card = 'Black'

How many members have the black membership card?

CREATE TABLE member (Membership_card VARCHAR)

Quantos membros têm o cartão de membro preto?

SELECT COUNT(*), address FROM member GROUP BY address

Find the number of members living in each address.

CREATE TABLE member (address VARCHAR)

Encontre o número de membros que vivem em cada endereço.

SELECT name FROM member WHERE address = 'Harford' OR address = 'Waterbury'

Give me the names of members whose address is in Harford or Waterbury.

CREATE TABLE member (name VARCHAR, address VARCHAR)

Dê-me os nomes dos membros cujo endereço é em Harford ou Waterbury.

SELECT name, member_id FROM member WHERE Membership_card = 'Black' OR age < 30

Find the ids and names of members who are under age 30 or with black membership card.

CREATE TABLE member (name VARCHAR, member_id VARCHAR, Membership_card VARCHAR, age VARCHAR)

Encontre os IDs e nomes dos membros com menos de 30 anos ou com cartão de membro preto.

SELECT Time_of_purchase, age, address FROM member ORDER BY Time_of_purchase

Find the purchase time, age and address of each member, and show the results in the order of purchase time.

CREATE TABLE member (Time_of_purchase VARCHAR, age VARCHAR, address VARCHAR)

Encontre o tempo de compra, a idade e o endereço de cada membro e mostre os resultados na ordem do tempo de compra.

SELECT Membership_card FROM member GROUP BY Membership_card HAVING COUNT(*) > 5

Which membership card has more than 5 members?

CREATE TABLE member (Membership_card VARCHAR)

Qual cartão de membro tem mais de 5 membros?

SELECT address FROM member WHERE age < 30 INTERSECT SELECT address FROM member WHERE age > 40

Which address has both members younger than 30 and members older than 40?

CREATE TABLE member (address VARCHAR, age INTEGER)

Qual endereço tem ambos os membros com menos de 30 anos e membros com mais de 40 anos?

SELECT membership_card FROM member WHERE address = 'Hartford' INTERSECT SELECT membership_card FROM member WHERE address = 'Waterbury'

What is the membership card held by both members living in Hartford and ones living in Waterbury address?

CREATE TABLE member (membership_card VARCHAR, address VARCHAR)

Qual é o cartão de membro mantido por ambos os membros que vivem em Hartford e aqueles que vivem no endereço de Waterbury?

SELECT COUNT(*) FROM member WHERE address <> 'Hartford'

How many members are not living in Hartford?

CREATE TABLE member (address VARCHAR)

Quantos membros não vivem em Hartford?

SELECT address FROM member EXCEPT SELECT address FROM member WHERE Membership_card = 'Black'

Which address do not have any member with the black membership card?

CREATE TABLE member (address VARCHAR, Membership_card VARCHAR)

Qual endereço não tem nenhum membro com o cartão de membro preto?

SELECT address FROM shop ORDER BY open_year

Show the shop addresses ordered by their opening year.

CREATE TABLE shop (address VARCHAR, open_year VARCHAR)

Mostre os endereços da loja encomendados pelo ano de abertura.

SELECT AVG(num_of_staff), AVG(score) FROM shop

What are the average score and average staff number of all shops?

CREATE TABLE shop (num_of_staff INTEGER, score INTEGER)

Qual é a pontuação média e o número médio de funcionários de todas as lojas?

SELECT shop_id, address FROM shop WHERE score < (SELECT AVG(score) FROM shop)

Find the id and address of the shops whose score is below the average score.

CREATE TABLE shop (shop_id VARCHAR, address VARCHAR, score INTEGER)

Encontre o ID e o endereço das lojas cuja pontuação está abaixo da pontuação média.

SELECT address, num_of_staff FROM shop WHERE NOT shop_id IN (SELECT shop_id FROM happy_hour)

Find the address and staff number of the shops that do not have any happy hour.

CREATE TABLE shop (address VARCHAR, num_of_staff VARCHAR, shop_id VARCHAR); CREATE TABLE happy_hour (address VARCHAR, num_of_staff VARCHAR, shop_id VARCHAR)

Encontre o endereço e o número de funcionários das lojas que não têm nenhum happy hour.

SELECT t1.address, t1.shop_id FROM shop AS t1 JOIN happy_hour AS t2 ON t1.shop_id = t2.shop_id WHERE MONTH = 'May'

What are the id and address of the shops which have a happy hour in May?

CREATE TABLE shop (address VARCHAR, shop_id VARCHAR); CREATE TABLE happy_hour (shop_id VARCHAR)

Quais são o ID e endereço das lojas que têm um happy hour em maio?

SELECT shop_id, COUNT(*) FROM happy_hour GROUP BY shop_id ORDER BY COUNT(*) DESC LIMIT 1

which shop has happy hour most frequently? List its id and number of happy hours.

CREATE TABLE happy_hour (shop_id VARCHAR)

qual loja tem happy hour mais frequentemente? Liste seu id e número de happy hours.

SELECT MONTH FROM happy_hour GROUP BY MONTH ORDER BY COUNT(*) DESC LIMIT 1

Which month has the most happy hours?

CREATE TABLE happy_hour (MONTH VARCHAR)

Qual mês tem mais horas felizes?

SELECT MONTH FROM happy_hour GROUP BY MONTH HAVING COUNT(*) > 2

Which months have more than 2 happy hours?

CREATE TABLE happy_hour (MONTH VARCHAR)

Quais meses têm mais de 2 horas felizes?

SELECT COUNT(*) FROM ALBUM

How many albums are there?

CREATE TABLE ALBUM (Id VARCHAR)

Quantos álbuns existem?

SELECT Name FROM GENRE

List the names of all music genres.

CREATE TABLE GENRE (Name VARCHAR)

Listar os nomes de todos os gêneros musicais.

SELECT * FROM CUSTOMER WHERE State = "NY"

Find all the customer information in state NY.

CREATE TABLE CUSTOMER (State VARCHAR)

Encontre todas as informações do cliente no estado de Nova York.

SELECT FirstName, LastName FROM EMPLOYEE WHERE City = "Calgary"

What are the first names and last names of the employees who live in Calgary city.

CREATE TABLE EMPLOYEE (FirstName VARCHAR, LastName VARCHAR, City VARCHAR)

Quais são os primeiros nomes e sobrenomes dos funcionários que vivem na cidade de Calgary?

SELECT DISTINCT (BillingCountry) FROM INVOICE

What are the distinct billing countries of the invoices?

CREATE TABLE INVOICE (BillingCountry VARCHAR)

Quais são os diferentes países de faturamento das faturas?

SELECT Name FROM ARTIST WHERE Name LIKE "%a%"

Find the names of all artists that have "a" in their names.

CREATE TABLE ARTIST (Name VARCHAR)

Encontre os nomes de todos os artistas que têm "a" em seus nomes.

SELECT Title FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "AC/DC"

Find the title of all the albums of the artist "AC/DC".

CREATE TABLE ARTIST (ArtistId VARCHAR, Name VARCHAR); CREATE TABLE ALBUM (ArtistId VARCHAR)

Encontre o título de todos os álbuns do artista "AC/DC".

SELECT COUNT(*) FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T2.Name = "Metallica"

Hom many albums does the artist "Metallica" have?

CREATE TABLE ARTIST (ArtistId VARCHAR, Name VARCHAR); CREATE TABLE ALBUM (ArtistId VARCHAR)

Hom muitos álbuns o artista "Metallica" tem?

SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId WHERE T1.Title = "Balls to the Wall"

Which artist does the album "Balls to the Wall" belong to?

CREATE TABLE ALBUM (ArtistId VARCHAR, Title VARCHAR); CREATE TABLE ARTIST (Name VARCHAR, ArtistId VARCHAR)

A qual artista pertence o álbum "Balls to the Wall"?

SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId GROUP BY T2.Name ORDER BY COUNT(*) DESC LIMIT 1

Which artist has the most albums?

CREATE TABLE ALBUM (ArtistId VARCHAR); CREATE TABLE ARTIST (Name VARCHAR, ArtistId VARCHAR)

Qual artista tem mais álbuns?

SELECT Name FROM TRACK WHERE Name LIKE '%you%'

Find the names of all the tracks that contain the word "you".

CREATE TABLE TRACK (Name VARCHAR)

Encontre os nomes de todas as faixas que contêm a palavra "você".

SELECT AVG(UnitPrice) FROM TRACK

What is the average unit price of all the tracks?

CREATE TABLE TRACK (UnitPrice INTEGER)

Qual é o preço unitário médio de todas as faixas?

SELECT MAX(Milliseconds), MIN(Milliseconds) FROM TRACK

What are the durations of the longest and the shortest tracks in milliseconds?

CREATE TABLE TRACK (Milliseconds INTEGER)

Quais são as durações das faixas mais longas e mais curtas em milissegundos?

SELECT T1.Title, T2.AlbumID, COUNT(*) FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId GROUP BY T2.AlbumID

Show the album names, ids and the number of tracks for each album.

CREATE TABLE ALBUM (Title VARCHAR, AlbumId VARCHAR); CREATE TABLE TRACK (AlbumID VARCHAR, AlbumId VARCHAR)

Mostre os nomes do álbum, IDs e o número de faixas para cada álbum.

SELECT T1.Name FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId GROUP BY T2.GenreId ORDER BY COUNT(*) DESC LIMIT 1

What is the name of the most common genre in all tracks?

CREATE TABLE GENRE (Name VARCHAR, GenreId VARCHAR); CREATE TABLE TRACK (GenreId VARCHAR)

Qual é o nome do gênero mais comum em todas as faixas?

SELECT T1.Name FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId GROUP BY T2.MediaTypeId ORDER BY COUNT(*) LIMIT 1

What is the least common media type in all tracks?

CREATE TABLE MEDIATYPE (Name VARCHAR, MediaTypeId VARCHAR); CREATE TABLE TRACK (MediaTypeId VARCHAR)

Qual é o tipo de mídia menos comum em todas as faixas?

SELECT T1.Title, T2.AlbumID FROM ALBUM AS T1 JOIN TRACK AS T2 ON T1.AlbumId = T2.AlbumId WHERE T2.UnitPrice > 1 GROUP BY T2.AlbumID

Show the album names and ids for albums that contain tracks with unit price bigger than 1.

CREATE TABLE ALBUM (Title VARCHAR, AlbumId VARCHAR); CREATE TABLE TRACK (AlbumID VARCHAR, AlbumId VARCHAR, UnitPrice INTEGER)

Mostre os nomes e IDs dos álbuns que contêm faixas com preço unitário maior que 1.

SELECT COUNT(*) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock"

How many tracks belong to rock genre?

CREATE TABLE TRACK (GenreId VARCHAR); CREATE TABLE GENRE (GenreId VARCHAR, Name VARCHAR)

Quantas faixas pertencem ao gênero rock?

SELECT AVG(UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Jazz"

What is the average unit price of tracks that belong to Jazz genre?

CREATE TABLE TRACK (GenreId VARCHAR); CREATE TABLE GENRE (GenreId VARCHAR, Name VARCHAR)

Qual é o preço unitário médio das faixas que pertencem ao gênero Jazz?

SELECT FirstName, LastName FROM CUSTOMER WHERE Email = "luisg@embraer.com.br"

What is the first name and last name of the customer that has email "luisg@embraer.com.br"?

CREATE TABLE CUSTOMER (FirstName VARCHAR, LastName VARCHAR, Email VARCHAR)

Qual é o primeiro nome e sobrenome do cliente que tem o e-mail "luisgembraer.com.br"?

SELECT COUNT(*) FROM CUSTOMER WHERE Email LIKE "%gmail.com%"

How many customers have email that contains "gmail.com"?

CREATE TABLE CUSTOMER (Email VARCHAR)

Quantos clientes têm e-mail que contém "gmail.com"?

SELECT T2.FirstName, T2.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.FirstName = "Leonie"

What is the first name and last name employee helps the customer with first name Leonie?

CREATE TABLE CUSTOMER (SupportRepId VARCHAR, FirstName VARCHAR); CREATE TABLE EMPLOYEE (FirstName VARCHAR, LastName VARCHAR, EmployeeId VARCHAR)

Qual é o primeiro nome e último nome funcionário ajuda o cliente com o nome próprio?

SELECT T2.City FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId WHERE T1.PostalCode = "70174"

What city does the employee who helps the customer with postal code 70174 live in?

CREATE TABLE EMPLOYEE (City VARCHAR, EmployeeId VARCHAR); CREATE TABLE CUSTOMER (SupportRepId VARCHAR, PostalCode VARCHAR)

Em que cidade vive o funcionário que ajuda o cliente com o código postal 70174?

SELECT COUNT(DISTINCT city) FROM EMPLOYEE

How many distinct cities does the employees live in?

CREATE TABLE EMPLOYEE (city VARCHAR)

Em quantas cidades os funcionários vivem?

SELECT T2.InvoiceDate FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.FirstName = "Astrid" AND LastName = "Gruber"

Find all invoice dates corresponding to customers with first name Astrid and last name Gruber.

CREATE TABLE CUSTOMER (CustomerId VARCHAR, FirstName VARCHAR); CREATE TABLE INVOICE (InvoiceDate VARCHAR, CustomerId VARCHAR)

Encontre todas as datas da fatura correspondentes a clientes com o primeiro nome Astrid e o último nome Gruber.

SELECT LastName FROM CUSTOMER EXCEPT SELECT T1.LastName FROM CUSTOMER AS T1 JOIN Invoice AS T2 ON T1.CustomerId = T2.CustomerId WHERE T2.total > 20

Find all the customer last names that do not have invoice totals larger than 20.

CREATE TABLE CUSTOMER (LastName VARCHAR); CREATE TABLE CUSTOMER (LastName VARCHAR, CustomerId VARCHAR); CREATE TABLE Invoice (CustomerId VARCHAR, total INTEGER)

Encontre todos os sobrenomes de clientes que não têm totais de fatura maiores que 20.

SELECT DISTINCT T1.FirstName FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Brazil"

Find the first names of all customers that live in Brazil and have an invoice.

CREATE TABLE CUSTOMER (FirstName VARCHAR, CustomerId VARCHAR, country VARCHAR); CREATE TABLE INVOICE (CustomerId VARCHAR)

Encontre os primeiros nomes de todos os clientes que vivem no Brasil e têm uma fatura.

SELECT DISTINCT T1.Address FROM CUSTOMER AS T1 JOIN INVOICE AS T2 ON T1.CustomerId = T2.CustomerId WHERE T1.country = "Germany"

Find the address of all customers that live in Germany and have invoice.

CREATE TABLE INVOICE (CustomerId VARCHAR); CREATE TABLE CUSTOMER (Address VARCHAR, CustomerId VARCHAR, country VARCHAR)

Encontre o endereço de todos os clientes que vivem na Alemanha e têm fatura.

SELECT Phone FROM EMPLOYEE

List the phone numbers of all employees.

CREATE TABLE EMPLOYEE (Phone VARCHAR)

Liste os números de telefone de todos os funcionários.

SELECT COUNT(*) FROM MEDIATYPE AS T1 JOIN TRACK AS T2 ON T1.MediaTypeId = T2.MediaTypeId WHERE T1.Name = "AAC audio file"

How many tracks are in the AAC audio file media type?

CREATE TABLE MEDIATYPE (MediaTypeId VARCHAR, Name VARCHAR); CREATE TABLE TRACK (MediaTypeId VARCHAR)

Quantas faixas estão no tipo de mídia de arquivo de áudio AAC?

SELECT AVG(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Latin" OR T1.Name = "Pop"

What is the average duration in milliseconds of tracks that belong to Latin or Pop genre?

CREATE TABLE TRACK (GenreId VARCHAR); CREATE TABLE GENRE (GenreId VARCHAR, Name VARCHAR)

Qual é a duração média em milissegundos de faixas que pertencem ao gênero latino ou pop?

SELECT T1.FirstName, T1.SupportRepId FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) >= 10

Please show the employee first names and ids of employees who serve at least 10 customers.

CREATE TABLE CUSTOMER (FirstName VARCHAR, SupportRepId VARCHAR); CREATE TABLE EMPLOYEE (EmployeeId VARCHAR)

Por favor, mostre os primeiros nomes e IDs dos funcionários que atendem pelo menos 10 clientes.

SELECT T1.LastName FROM CUSTOMER AS T1 JOIN EMPLOYEE AS T2 ON T1.SupportRepId = T2.EmployeeId GROUP BY T1.SupportRepId HAVING COUNT(*) <= 20

Please show the employee last names that serves no more than 20 customers.

CREATE TABLE EMPLOYEE (EmployeeId VARCHAR); CREATE TABLE CUSTOMER (LastName VARCHAR, SupportRepId VARCHAR)

Por favor, mostre os sobrenomes dos funcionários que não atendem mais de 20 clientes.

SELECT Title FROM ALBUM ORDER BY Title

Please list all album titles in alphabetical order.

CREATE TABLE ALBUM (Title VARCHAR)

Por favor, liste todos os títulos do álbum em ordem alfabética.

SELECT T2.Name, T1.ArtistId FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistID GROUP BY T1.ArtistId HAVING COUNT(*) >= 3 ORDER BY T2.Name

Please list the name and id of all artists that have at least 3 albums in alphabetical order.

CREATE TABLE ARTIST (Name VARCHAR, ArtistID VARCHAR); CREATE TABLE ALBUM (ArtistId VARCHAR)

Por favor, liste o nome e o ID de todos os artistas que tenham pelo menos 3 álbuns em ordem alfabética.

SELECT Name FROM ARTIST EXCEPT SELECT T2.Name FROM ALBUM AS T1 JOIN ARTIST AS T2 ON T1.ArtistId = T2.ArtistId

Find the names of artists that do not have any albums.

CREATE TABLE ARTIST (Name VARCHAR); CREATE TABLE ALBUM (ArtistId VARCHAR); CREATE TABLE ARTIST (Name VARCHAR, ArtistId VARCHAR)

Encontre os nomes dos artistas que não têm nenhum álbum.

SELECT AVG(T2.UnitPrice) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Rock"

What is the average unit price of rock tracks?

CREATE TABLE GENRE (GenreId VARCHAR, Name VARCHAR); CREATE TABLE TRACK (UnitPrice INTEGER, GenreId VARCHAR)

Qual é o preço unitário médio das faixas de rocha?

SELECT MAX(Milliseconds), MIN(Milliseconds) FROM GENRE AS T1 JOIN TRACK AS T2 ON T1.GenreId = T2.GenreId WHERE T1.Name = "Pop"

What are the duration of the longest and shortest pop tracks in milliseconds?

CREATE TABLE TRACK (GenreId VARCHAR); CREATE TABLE GENRE (GenreId VARCHAR, Name VARCHAR)

Qual é a duração das faixas pop mais longas e curtas em milissegundos?

SELECT BirthDate FROM EMPLOYEE WHERE City = "Edmonton"

What are the birth dates of employees living in Edmonton?

CREATE TABLE EMPLOYEE (BirthDate VARCHAR, City VARCHAR)

Quais são as datas de nascimento dos funcionários que vivem em Edmonton?

SELECT DISTINCT (UnitPrice) FROM TRACK

What are the distinct unit prices of all tracks?

CREATE TABLE TRACK (UnitPrice VARCHAR)

Quais são os preços unitários distintos de todas as faixas?

SELECT COUNT(*) FROM ARTIST WHERE NOT artistid IN (SELECT artistid FROM ALBUM)

How many artists do not have any album?

CREATE TABLE ARTIST (artistid VARCHAR); CREATE TABLE ALBUM (artistid VARCHAR)

Quantos artistas não têm nenhum álbum?

SELECT T1.Title FROM Album AS T1 JOIN Track AS T2 ON T1.AlbumId = T2.AlbumId JOIN Genre AS T3 ON T2.GenreID = T3.GenreID WHERE T3.Name = 'Reggae' INTERSECT SELECT T1.Title FROM Album AS T1 JOIN Track AS T2 ON T1.AlbumId = T2.AlbumId JOIN Genre AS T3 ON T2.GenreID = T3.GenreID WHERE T3.Name = 'Rock'

What are the album titles for albums containing both 'Reggae' and 'Rock' genre tracks?

CREATE TABLE Genre (GenreID VARCHAR, Name VARCHAR); CREATE TABLE Track (AlbumId VARCHAR, GenreID VARCHAR); CREATE TABLE Album (Title VARCHAR, AlbumId VARCHAR)

Quais são os títulos do álbum para álbuns contendo faixas do gênero 'Reggae' e 'Rock'?

SELECT customer_phone FROM available_policies

Find all the phone numbers.

CREATE TABLE available_policies (customer_phone VARCHAR)

Encontre todos os números de telefone.

SELECT customer_phone FROM available_policies WHERE policy_type_code = "Life Insurance"

What are the customer phone numbers under the policy "Life Insurance"?

CREATE TABLE available_policies (customer_phone VARCHAR, policy_type_code VARCHAR)

Quais são os números de telefone do cliente sob a política "Seguro de Vida"?

SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY COUNT(*) DESC LIMIT 1

Which policy type has the most records in the database?

CREATE TABLE available_policies (policy_type_code VARCHAR)

Qual tipo de política tem mais registros no banco de dados?

SELECT customer_phone FROM available_policies WHERE policy_type_code = (SELECT policy_type_code FROM available_policies GROUP BY policy_type_code ORDER BY COUNT(*) DESC LIMIT 1)

What are all the customer phone numbers under the most popular policy type?

CREATE TABLE available_policies (customer_phone VARCHAR, policy_type_code VARCHAR)

Quais são todos os números de telefone do cliente sob o tipo de política mais popular?

SELECT policy_type_code FROM available_policies GROUP BY policy_type_code HAVING COUNT(*) > 4

Find the policy type used by more than 4 customers.

CREATE TABLE available_policies (policy_type_code VARCHAR)

Encontre o tipo de política usado por mais de 4 clientes.

SELECT SUM(settlement_amount), AVG(settlement_amount) FROM settlements

Find the total and average amount of settlements.

CREATE TABLE settlements (settlement_amount INTEGER)

Encontre a quantidade total e média de assentamentos.