Question

我总是看到COUNT（）与*一起用作参数而GROUP BY确定要计算的内容，但我相信你可以计算不同的东西，而不必进行子查询，分组，然后加入（这就是我现在正在做的事情）。

让我们用简化版的SO构建一个例子：

Table Question
idQuestion | title | idUser

Table Answer
idAnswer | idQUestion | text | idUser

Table Vote
idVote | idAnswer | type | idUser

Table Comment
idComment | idAnswer | text | idUser

要知道我所做的答案有多少评论和投票：

SELECT idAnswer, votes, comments FROM
   (SELECT idAnswer, count(*) AS votes FROM Answer A
       JOIN Vote B ON A.idAnswer=B.idAnswer
    GROUP BY idAnswer) AS X
JOIN
   (SELECT idAnswer, count(*) AS comments FROM Answer A
       JOIN Comment B ON A.idAnswer=B.idAnswer
    GROUP BY idAnswer) AS Y
ON X.idAnswer=Y.idAnswer

这可以在没有子选择的情况下完成吗？通过避免它们取得了什么成就？

如何在同一查询中计算负投票和正投票？

Answer 1

SELECT idAnswer, COUNT(DISTINCT VOTE.idVote), COUNT(DISTINCT comments)
FROM Answer 
LEFT JOIN COMMENT ON Answer.idAnswer = Comment.idAnswer
LEFT JOIN VOTE ON Answer.idAnswer = Vote.idAnswer
GROUP BY idAnswer

至于你问题的第二部分，感谢Andriy M！

SELECT
  Answer.idAnswer,
  COUNT(DISTINCT CASE WHEN VOTE.type = 'up' THEN VOTE.idVote END) AS upvotes,
  COUNT(DISTINCT CASE WHEN VOTE.type = 'down' THEN VOTE.idVote END) AS downvotes,
  COUNT(DISTINCT comments) as comments
FROM Answer 
LEFT JOIN COMMENT ON Answer.idAnswer = Comment.idAnswer
LEFT JOIN VOTE ON Answer.idAnswer = Vote.idAnswer
GROUP BY Answer.idAnswer

如何从一个大的JOIN表中计算多个东西

1 个答案: