Contino SQL Assessment

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SELECT
  STRING_AGG(t.TagName, ',') AS Tags,
  COUNT(t.TagName) AS NumberOfTags,
  p.Id AS PostId,
  p.OwnerUserId AS UserId,
  p.Body AS PostBody
FROM Tags AS t
INNER JOIN PostTags AS pt ON t.Id = pt.TagId
INNER JOIN Posts AS p ON pt.PostId = p.Id
GROUP BY p.Id, p.OwnerUserId, p.Body
HAVING COUNT(t.TagName) >= 5
ORDER BY COUNT(t.TagName) DESC;

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