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Databases Querying Relational Databases Subqueries Review and Practice

Posted by fairest
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fairest
Full Stack JavaScript Techdegree Graduate 19,303 Points

Is 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
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

A solution that works is a good solution. :wink:

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
seal-mask
.a{fill-rule:evenodd;}techdegree seal-36
fairest
Full Stack JavaScript Techdegree Graduate 19,303 Points
fairest
.a{fill-rule:evenodd;}techdegree seal-36
fairest
Full Stack JavaScript Techdegree Graduate 19,303 Points

Thanks Steven. Do you know what the error code would be?

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

It 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"