1

Please login or register to vote for this query.

(click on this box to dismiss)

Q&A about the site for linguists, teachers and students of the Korean language

SELECT
  RANK() OVER(ORDER BY COUNT(a.Id) DESC) AS [Rank],
  t.TagName,
  COUNT(a.Id) AS [Count],
  SUM(a.Score) AS [Score],
  SUM(case when (a.CommunityOwnedDate IS NULL) then 1 else 0 end) AS [non-CW],
  SUM(case when (a.CommunityOwnedDate IS NULL) then a.Score else 0 end) AS [non-CW score]
FROM Posts a INNER JOIN Posts q ON q.Id=a.ParentId
INNER JOIN PostTags pt ON pt.PostId=q.Id
INNER JOIN Tags t ON t.Id=pt.TagId
WHERE a.OwnerUserId=##userid?8297##
GROUP BY t.TagName
ORDER BY COUNT(a.Id) DESC

fork query

edited aug 10 20

created aug 9 20 Martin20528