[
    {
        "id": 1,
        "question": "List all the artists name in the database.",
        "taxonomy": "Selection",
        "ex_pass": true,
        "esm_pass": false,
        "generated_sql": "SELECT a.Name FROM Artist a;",
        "gold_sql": "SELECT Name FROM Artist;"
    },
    {
        "id": 2,
        "question": "How many genres are there?",
        "taxonomy": "Aggregation",
        "ex_pass": true,
        "esm_pass": true,
        "generated_sql": "SELECT COUNT(*) FROM Genre;",
        "gold_sql": "SELECT COUNT(*) FROM Genre;"
    },
    {
        "id": 3,
        "question": "List the names of the first 5 tracks.",
        "taxonomy": "Selection, Limit",
        "ex_pass": true,
        "esm_pass": false,
        "generated_sql": "SELECT t.Name FROM Track t ORDER BY t.TrackId LIMIT 5;",
        "gold_sql": "SELECT Name FROM Track LIMIT 5;"
    },
    {
        "id": 4,
        "question": "Count the number of customers located in the USA.",
        "taxonomy": "Aggregation, Filtering",
        "ex_pass": true,
        "esm_pass": false,
        "generated_sql": "SELECT COUNT(*) FROM Customer c WHERE c.Country = 'USA';",
        "gold_sql": "SELECT COUNT(*) FROM Customer WHERE Country = 'USA';"
    },
    {
        "id": 5,
        "question": "Find all invoices for the customer with ID 1.",
        "taxonomy": "Selection, Filtering",
        "ex_pass": false,
        "esm_pass": false,
        "generated_sql": "SELECT i.InvoiceId, i.InvoiceDate, i.Total FROM Invoice i WHERE i.CustomerId = 1;",
        "gold_sql": "SELECT * FROM Invoice WHERE CustomerId = 1;"
    },
    {
        "id": 6,
        "question": "List each album title along with the artist's name.",
        "taxonomy": "Simple Join",
        "ex_pass": true,
        "esm_pass": false,
        "generated_sql": "SELECT a.Title, ar.Name FROM Album a JOIN Artist ar ON a.ArtistId = ar.ArtistId;",
        "gold_sql": "SELECT Album.Title, Artist.Name FROM Album JOIN Artist ON Album.ArtistId = Artist.ArtistId;"
    },
    {
        "id": 7,
        "question": "How many tracks belong to the 'Rock' genre?",
        "taxonomy": "Simple Join, Filtering, Aggregation",
        "ex_pass": true,
        "esm_pass": false,
        "generated_sql": "SELECT COUNT(t.TrackId) FROM Track t JOIN Genre g ON t.GenreId = g.GenreId WHERE g.Name = 'Rock';",
        "gold_sql": "SELECT COUNT(*) FROM Track JOIN Genre ON Track.GenreId = Genre.GenreId WHERE Genre.Name = 'Rock';"
    },
    {
        "id": 8,
        "question": "Show the total revenue generated from each country.",
        "taxonomy": "Aggregation, Grouping",
        "ex_pass": true,
        "esm_pass": false,
        "generated_sql": "SELECT i.BillingCountry, SUM(i.Total) AS TotalRevenue FROM Invoice i GROUP BY i.BillingCountry ORDER BY TotalRevenue DESC NULLS LAST;",
        "gold_sql": "SELECT BillingCountry, SUM(Total) FROM Invoice GROUP BY BillingCountry;"
    },
    {
        "id": 9,
        "question": "Find the total number of items sold for each media type.",
        "taxonomy": "Multi-Join, Aggregation, Grouping",
        "ex_pass": false,
        "esm_pass": false,
        "generated_sql": "SELECT mt.Name, COUNT(*) AS total_items_sold FROM MediaType mt JOIN (SELECT MediaType.MediaTypeId, COUNT(*) AS items_sold FROM Album JOIN MediaType ON Album.MediaTypeId = MediaType.MediaTypeId GROUP BY MediaType.MediaTypeId) AS ms ON mt.MediaTypeId = ms.MediaTypeId GROUP BY mt.Name ORDER BY total_items_sold DESC NULLS LAST;",
        "gold_sql": "SELECT MediaType.Name, SUM(InvoiceLine.Quantity) FROM InvoiceLine JOIN Track ON InvoiceLine.TrackId = Track.TrackId JOIN MediaType ON Track.MediaTypeId = MediaType.MediaTypeId GROUP BY MediaType.Name;"
    },
    {
        "id": 10,
        "question": "List the first and last names of all employees who are Sales Support Agents.",
        "taxonomy": "Selection, Filtering",
        "ex_pass": true,
        "esm_pass": false,
        "generated_sql": "SELECT e.FirstName, e.LastName FROM Employee e WHERE e.Title = 'Sales Support Agent';",
        "gold_sql": "SELECT FirstName, LastName FROM Employee WHERE Title = 'Sales Support Agent';"
    },
    {
        "id": 11,
        "question": "List the top 5 customers who have spent the most money in total.",
        "taxonomy": "Simple Join, Aggregation, Grouping, Ordering, Limit",
        "ex_pass": false,
        "esm_pass": false,
        "generated_sql": "SELECT c.CustomerId, SUM(o.TotalAmount) AS TotalSpent FROM Customer c JOIN Order o ON c.CustomerId = o.CustomerId GROUP BY c.CustomerId ORDER BY TotalSpent DESC LIMIT 5;",
        "gold_sql": "SELECT c.FirstName, c.LastName, SUM(i.Total) as TotalSpent FROM Customer c JOIN Invoice i ON c.CustomerId = i.CustomerId GROUP BY c.CustomerId ORDER BY TotalSpent DESC LIMIT 5;"
    },
    {
        "id": 12,
        "question": "Which artist has the most tracks in the database? Give the name and count.",
        "taxonomy": "Multi-Join, Aggregation, Grouping, Ordering, Limit",
        "ex_pass": false,
        "esm_pass": false,
        "generated_sql": "SELECT a.\"ArtistId\", COUNT(a.\"AlbumId\") AS track_count FROM \"Album\" a GROUP BY a.\"ArtistId\" ORDER BY track_count DESC LIMIT 1;",
        "gold_sql": "SELECT ar.Name, COUNT(t.TrackId) as TrackCount FROM Artist ar JOIN Album al ON ar.ArtistId = al.ArtistId JOIN Track t ON al.AlbumId = t.AlbumId GROUP BY ar.ArtistId ORDER BY TrackCount DESC LIMIT 1;"
    },
    {
        "id": 13,
        "question": "Which genres have more than 100 tracks? List the genre name and count.",
        "taxonomy": "Simple Join, Aggregation, Grouping, Having",
        "ex_pass": true,
        "esm_pass": false,
        "generated_sql": "SELECT g.Name, COUNT(t.TrackId) AS TrackCount FROM Genre g JOIN Track t ON g.GenreId = t.GenreId GROUP BY g.Name HAVING COUNT(t.TrackId) > 100 ORDER BY TrackCount DESC NULLS LAST;",
        "gold_sql": "SELECT g.Name, COUNT(t.TrackId) as TrackCount FROM Genre g JOIN Track t ON g.GenreId = t.GenreId GROUP BY g.GenreId HAVING TrackCount > 100;"
    },
    {
        "id": 14,
        "question": "Calculate the average track length in seconds for each genre.",
        "taxonomy": "Simple Join, Aggregation, Arithmetic, Grouping",
        "ex_pass": false,
        "esm_pass": false,
        "generated_sql": "SELECT t.GenreId, AVG(t.Milliseconds / 1000.0) AS average_length FROM Track t GROUP BY t.GenreId ORDER BY t.GenreId NULLS LAST;",
        "gold_sql": "SELECT g.Name, AVG(t.Milliseconds) / 1000.0 as AvgSeconds FROM Genre g JOIN Track t ON g.GenreId = t.GenreId GROUP BY g.GenreId;"
    },
    {
        "id": 15,
        "question": "Identify the artist who has earned the most revenue from customers in Canada.",
        "taxonomy": "Multi-Join, Aggregation, Grouping, Ordering, Limit",
        "ex_pass": false,
        "esm_pass": false,
        "generated_sql": "SELECT a.Name, SUM(al.Sales) AS TotalRevenue FROM Artist a JOIN Album al ON a.ArtistId = al.ArtistId JOIN Customer c ON al.CustomerId = c.CustomerId WHERE c.Country = 'Canada' GROUP BY a.Name ORDER BY TotalRevenue DESC LIMIT 1;",
        "gold_sql": "SELECT ar.Name, SUM(il.UnitPrice * il.Quantity) AS Revenue FROM Artist ar JOIN Album al ON ar.ArtistId = al.ArtistId JOIN Track t ON al.AlbumId = t.AlbumId JOIN InvoiceLine il ON t.TrackId = il.TrackId JOIN Invoice i ON il.InvoiceId = i.InvoiceId WHERE i.BillingCountry = 'Canada' GROUP BY ar.ArtistId ORDER BY Revenue DESC LIMIT 1;"
    }
]