Chinook Database SQL

Chinook is a sample database available for SQL Server, Oracle, MySQL, etc. It can be created by running a single SQL script. Chinook database is an alternative to the Northwind database, being ideal for demos and testing ORM tools targeting single and multiple database servers.

Specifying SQL Queries:

1. Provide a query showing Customers (just their full names, customer ID and country) who are not in the US.

SELECT
  "CustomerId",
  "FirstName",
  "LastName",
  "Country"
FROM
  "Customer"
WHERE
  "Country" <> 'USA';

2. Provide a query only showing the Customers from Brazil.

SELECT
  "CustomerId",
  "FirstName",
  "LastName",
  "Country"
FROM
  "Customer"
WHERE
  "Country" = 'Brazil';

3. Provide a query showing the Invoices of customers who are from Brazil. The resultant table should show the customer's full name, Invoice ID, Date of the invoice and billing country.

SELECT
  "Customer"."CustomerId",
  "Customer"."FirstName" || ' ' || "Customer"."LastName" AS "Customer",
  "Invoice"."InvoiceId",
  "Invoice"."InvoiceDate",
  "Invoice"."BillingCountry"
FROM
  "Customer"
  JOIN "Invoice" USING("CustomerId")
WHERE
  "Invoice"."BillingCountry" = 'USA';

4. Provide a query showing only the Employees who are Sales Agents.

SELECT
  *
FROM
  "Employee";

5. Provide a query showing a unique list of billing countries from the Invoice table.

SELECT 
	DISTINCT "Invoice"."BillingCountry"
FROM
  "Invoice";

6. Provide a query that shows the invoices associated with each sales agent. The resultant table should include the Sales Agent's full name.

SELECT 
	"Employee"."LastName" || ' ' || "Employee"."FirstName" AS "Employee",
	"Invoice".*
FROM 
	"Invoice"
  	JOIN "Customer" USING("CustomerId")
	JOIN "Employee" ON "Customer"."SupportRepId" = "Employee"."EmployeeId";

7. Provide a query that shows the Invoice Total, Customer name, Country and Sale Agent name for all invoices and customers.

SELECT
  	"Customer"."FirstName" || ' ' || "Customer"."LastName" AS "Customer",
  	"Invoice"."BillingCountry",
  	"Invoice"."Total",
  	"Employee"."LastName"
FROM
  	"Customer"
  	JOIN "Invoice" USING("CustomerId")
 	JOIN "Employee" ON "Customer"."SupportRepId" = "Employee"."EmployeeId";

8. How many Invoices were there in 2009 and 2011? What are the respective total sales for each of those years?

SELECT 
	COUNT("Invoice"."InvoiceId"),
	SUM("Invoice"."Total") AS "Total"
FROM
	"Invoice"
WHERE
	"Invoice"."InvoiceDate" BETWEEN '2009-01-01' AND '2009-12-31' OR 
	"Invoice"."InvoiceDate" BETWEEN '2011-01-01' AND '2011-12-31'

9. Looking at the InvoiceLine table, provide a query that COUNTs the number of line items for Invoice ID 37.

SELECT 
	COUNT("InvoiceLine"."InvoiceLineId") AS "NumberOfLineItems"
FROM 
	"InvoiceLine"
WHERE 
	"InvoiceLine"."InvoiceLineId" = 37;

10. Looking at the InvoiceLine table, provide a query that COUNTs the number of line items for each Invoice. HINT: GROUP BY

SELECT
	"InvoiceLine"."InvoiceId" AS "InvoiceId",
	COUNT("InvoiceLine"."InvoiceId")
FROM
	"InvoiceLine"
	GROUP BY("InvoiceId");

11. Provide a query that includes the track name with each invoice line item.

SELECT 
	"Track"."Name" AS "trackName",
	"InvoiceLine".*
FROM
	"Track"
	JOIN "InvoiceLine" USING("TrackId");

12. Provide a query that includes the purchased track name AND artist name with each invoice line item.

SELECT 
	"Artist"."Name" AS "NameArtist",
	"Track"."Composer",
	"Track"."Name" AS "NameTrack",
	"InvoiceLine"."InvoiceId"
FROM 
	"Artist"
	JOIN "Album" USING("ArtistId")
	JOIN "Track" USING("AlbumId")
	JOIN "InvoiceLine" USING("TrackId");

13. Provide a query that shows the # of invoices per country. HINT: GROUP BY

SELECT 
	"Invoice"."BillingCountry", 
	COUNT(*) AS "Number_of_Invoices"
FROM 
	"Invoice"
GROUP BY ("Invoice"."BillingCountry")
ORDER BY ("Number_of_Invoices") ASC;

14. Provide a query that shows the total number of tracks in each playlist. The Playlist name should be included on the resultant table.

SELECT 
	"Playlist"."Name" AS "Name_Playlist", 
	COUNT(*) AS "Number_of_Tracks"
FROM 
	"PlaylistTrack"
	JOIN "Playlist" USING("PlaylistId")
	JOIN "Track" USING("TrackId")
	JOIN "InvoiceLine" USING("TrackId")
	JOIN "Invoice" USING("InvoiceId")
GROUP BY ("Playlist"."Name")
ORDER BY ("Number_of_Tracks") ASC;

15. Provide a query that shows all the Tracks, but displays no IDs. The resultant table should include the Album name, Media type and Genre.

SELECT 
	"Track"."Name" AS "TrackName",
	"Album"."Title" AS "AlbumTitle",
	"Genre"."Name" AS "Genre",
	"MediaTypeId" AS "MediaType"
FROM 
	"Track"
	JOIN "Album" USING("AlbumId")
	JOIN "Genre" USING("GenreId")
	JOIN "MediaType" USING("MediaTypeId");

16. Provide a query that shows all Invoices but includes the # of invoice line items.

SELECT 
	"Invoice".*, 
	COUNT("InvoiceLine"."InvoiceId") AS "ofLineItems"
FROM 
	"InvoiceLine"
	JOIN "Invoice" USING("InvoiceId")
GROUP BY ("Invoice"."InvoiceId");

17. Provide a query that shows total sales made by each sales agent.

SELECT 
	DISTINCT "Employee"."LastName" || ' ' || "Employee"."FirstName" AS "SalesAgent",
	SUM("Invoice"."Total") AS "Total"
FROM
	"Customer"
	JOIN "Employee" ON "Customer"."SupportRepId" = "Employee"."EmployeeId"
	JOIN "Invoice" USING("CustomerId")
GROUP BY ("SalesAgent")
ORDER BY ("Total") DESC;

18. Which sales agent made the most in sales in 2009?

SELECT 
	DISTINCT "Employee"."LastName" || ' ' || "Employee"."FirstName" AS "SalesAgent",
	SUM("Invoice"."Total") AS "Total"
FROM
	"Customer"
	JOIN "Employee" ON "Customer"."SupportRepId" = "Employee"."EmployeeId"
	JOIN "Invoice" USING("CustomerId")
WHERE 
	"Invoice"."InvoiceDate" BETWEEN '2009-01-01' AND '2009-12-31'
GROUP BY ("SalesAgent")
ORDER BY ("Total") DESC
LIMIT 1;

19. Which sales agent made the most in sales in 2010?

SELECT 
	DISTINCT "Employee"."LastName" || ' ' || "Employee"."FirstName" AS "SalesAgent",
	SUM("Invoice"."Total") AS "Total"
FROM
	"Customer"
	JOIN "Employee" ON "Customer"."SupportRepId" = "Employee"."EmployeeId"
	JOIN "Invoice" USING("CustomerId")
WHERE 
	"Invoice"."InvoiceDate" BETWEEN '2010-01-01' AND '2010-12-31'
GROUP BY ("SalesAgent")
ORDER BY ("Total") DESC
LIMIT 1;

20. Which sales agent made the most in sales over all?

SELECT 
	DISTINCT "Employee"."LastName" || ' ' || "Employee"."FirstName" AS "SalesAgent",
	SUM("Invoice"."Total") AS "Total"
FROM
	"Customer"
	JOIN "Employee" ON "Customer"."SupportRepId" = "Employee"."EmployeeId"
	JOIN "Invoice" USING("CustomerId")
GROUP BY ("SalesAgent")
ORDER BY ("Total") DESC
LIMIT 1;

21. Provide a query that shows the # of customers assigned to each sales agent.

SELECT 
	"Employee"."LastName" || ' ' || "Employee"."FirstName" AS "SalesAgent",
	COUNT("Customer"."SupportRepId") AS "TopSalesAgent"
FROM
	"Customer"
	JOIN "Employee" ON "Customer"."SupportRepId" = "Employee"."EmployeeId"
	JOIN "Invoice" USING("CustomerId")
GROUP BY ("SalesAgent")
ORDER BY ("TopSalesAgent") DESC
LIMIT 1;

22. Provide a query that shows the total sales per country. Which country's customers spent the most?

SELECT
	DISTINCT "Invoice"."BillingCountry" AS "Country",
	SUM("Invoice"."Total") AS "Total"
FROM 
	"Invoice"
GROUP BY ("Country")
ORDER BY ("Total") DESC
LIMIT 1;

23. Provide a query that shows the most purchased track of 2013.

SELECT
	"Track"."Composer" AS "ComposerName",
	COUNT("InvoiceLine"."InvoiceLineId") AS "topComposer"
FROM 
	"Track"
	JOIN "InvoiceLine" USING("TrackId")
	JOIN "Invoice" USING("InvoiceId")
WHERE 
	"Invoice"."InvoiceDate" BETWEEN  '2013-01-01' AND '2013-12-31'
GROUP BY ("ComposerName")
ORDER BY ("topComposer") DESC
LIMIT 1;

24. Provide a query that shows the top 5 most purchased tracks over all.

SELECT 
	"Track"."Name",
	COUNT("InvoiceLine"."InvoiceLineId") AS "topTrack"
FROM 
	"Track"
	JOIN "InvoiceLine" USING("TrackId")
GROUP BY ("Track"."Name")
ORDER BY ("topTrack") DESC
LIMIT 5;

25. Provide a query that shows the top 3 best selling artists.

SELECT 
	"Artist"."Name" AS "Artist",
	COUNT("InvoiceLine"."InvoiceLineId") AS "topArtist"
FROM 
	"Album"
	JOIN "Artist" USING("ArtistId")
	JOIN "Track" USING("AlbumId")
	JOIN "InvoiceLine" USING("TrackId")
GROUP BY ("Artist")
ORDER BY ("topArtist") DESC
LIMIT 3;

26. Provide a query that shows the most purchased Media Type.

SELECT 
	"MediaType"."Name" AS "MediaType",
	COUNT("InvoiceLine"."InvoiceLineId") AS "topMediaType"
FROM 
	"Track"
	JOIN "MediaType" USING("MediaTypeId")
	JOIN "InvoiceLine" USING("TrackId")
GROUP BY ("MediaType")
ORDER BY ("topMediaType") DESC
LIMIT 1;

27. Provide a query that shows the number tracks purchased in all invoices that contain more than one genre.

SELECT
	DISTINCT "MediaType"."Name",
	COUNT("Track"."TrackId") AS "PurchasedСompositions"
FROM 
	"Track"
	JOIN "MediaType" USING("MediaTypeId")
	JOIN "Genre" USING("GenreId")
	JOIN "InvoiceLine" USING("TrackId")
	JOIN "Invoice" USING("InvoiceId")
WHERE
	"Track"."GenreId" > 1
GROUP BY ("MediaType"."Name")
ORDER BY ("PurchasedСompositions") DESC;

EMD SCHEME: