SQL_data_science_quiz_3.sql

--Q1) Using a subquery, find the names of all the tracks for the
-- album "Californication".

SELECT Name,
       AlbumID
FROM Tracks
WHERE AlbumId IN (SELECT AlbumId
    FROM Albums
    WHERE Title = "Californication");
    
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-- Q2) Find the total number of invoices for each customer along 
-- with the customer's full name, city and email.

SELECT FirstName,
       LastName,
       City,
       Email,
       COUNT(I.CustomerId) AS Invoices
FROM Customers C INNER JOIN Invoices I
ON C.CustomerId = I.CustomerId
GROUP BY C.CustomerId

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-- Q3) Retrieve the track name, album, artist, and trackID for 
-- all the albums.

SELECT Tracks.Name,
       A.Name AS Artist,
       Albums.Title AS Album,
       Tracks.TrackId
FROM ((Tracks INNER JOIN Albums
ON Tracks.AlbumId = Albums.AlbumId)
INNER JOIN Artists A
ON A.ArtistId = Albums.ArtistId); 

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-- Q4) Retrieve a list with the managers last name, and the last 
-- name of the employees who report to him or her.

SELECT M.LastName AS Manager, 
       E.LastName AS Employee
FROM Employees E INNER JOIN Employees M 
ON E.ReportsTo = M.EmployeeID

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-- Q5) Find the name and ID of the artists who do not have albums.

SELECT Name AS Artist,
       Artists.ArtistId,
       Albums.Title AS Album
FROM Artists
LEFT JOIN Albums
ON Artists.ArtistId = Albums.ArtistId
WHERE Album IS NULL

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-- Q6) Use a UNION to create a list of all the employee's & 
-- customer's first names and last names ordered by the last
-- name in descending order.

SELECT FirstName,
       LastName
FROM Employees
UNION
SELECT FirstName,
       LastName
FROM Customers
ORDER BY LastName DESC

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-- Q7) See if there are any customers who have a different city
-- listed in their billing city versus their customer city.

SELECT C.FirstName,
       C.LastName,
       C.City AS CustomerCity,
       I.BillingCity
FROM Customers C
INNER JOIN Invoices I
ON C.CustomerId = I.CustomerId
WHERE CustomerCity != BillingCity