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fairest
Full Stack JavaScript Techdegree Graduate 19,303 PointsIs this a decent solution for challenge 2?
SELECT p.first_name, p.email, COUNT(p.id) AS Outstanding FROM patrons AS p
INNER JOIN (
SELECT * FROM loans_north
UNION ALL
SELECT * FROM loans_south
) AS l ON p.id = l.patron_id
WHERE l.returned_on IS NULL
GROUP BY p.id
ORDER BY p.first_name;
2 Answers
Steven Parker
231,184 PointsA solution that works is a good solution.
However, it's "best practice" to always include any SELECTed column that isn't in an aggregating function in the GROUP BY clause. Some database engines will give you an error if you don't.
fairest
Full Stack JavaScript Techdegree Graduate 19,303 PointsThanks Steven. Do you know what the error code would be?
Steven Parker
231,184 PointsIt depends on the specific database engine, but a typical example message might be:
"ERROR: column "<column_name>" must appear in the GROUP BY clause or be used in an aggregate function"