Question

我正在撰写a data explorer query以找到得分最低的答案，这部分工作正常：

SELECT TOP 25 a.Id as [Post Link],
              a.Score as Score,
              (SELECT COUNT(*)
               FROM Votes
               WHERE PostId = a.Id AND
                     VoteTypeId = 2) AS [Upvotes],
              (SELECT COUNT(*)
               FROM Votes
               WHERE PostId = a.Id AND
                     VoteTypeId = 3) AS [Downvotes]
FROM Posts q INNER JOIN Posts a ON q.AcceptedAnswerId = a.Id
WHERE a.PostTypeId = 2
ORDER BY a.Score ASC

但是，当我尝试找出发布答案的人得分时，我无法接受我选择的[Upvotes]和[Downvotes]列作为参数，即：

SELECT TOP 25 a.Id as [Post Link],
              a.Score as Score,
              (SELECT COUNT(*)
               FROM Votes
               WHERE PostId = a.Id AND
                     VoteTypeId = 2) AS [Upvotes],
              (SELECT COUNT(*)
               FROM Votes
               WHERE PostId = a.Id AND
                     VoteTypeId = 3) AS [Downvotes],
              (15 + (([Upvotes] * 10) - ([Downvotes] * 2))) AS [Answerer's Reputation]
FROM Posts q INNER JOIN Posts a ON q.AcceptedAnswerId = a.Id
WHERE a.PostTypeId = 2
ORDER BY a.Score ASC

我告诉[Upvotes]，[Downvotes]不是有效的列名：

列名称无效＆＃39; Upvotes＆＃39;。列名称无效＆＃39; Downvotes＆＃39;。

如何表明[Upvotes]与我刚宣布的AS [Upvotes]相同？{{1}}？或者这根本不可能，我必须手动将其作为存储过程吗？

Answer 1

问题是计算列是在引用列的同时计算的，因此SQL不了解它们。

这将是一个解决方案

SELECT A.[Post Link], 
   A.Score, 
   A.Upvotes, 
   A.Downvotes, 
   (15 + (([Upvotes] * 10) - ([Downvotes] * 2)))AS [Answerer's Reputation]
FROM(SELECT TOP 25 P.Id AS [Post Link], 
                 P.Score AS Score, 
                 (
   SELECT COUNT(*)
     FROM Votes
    WHERE Postid = P.Id
      AND Votetypeid = 2)AS [Upvotes], 
                 (
   SELECT COUNT(*)
     FROM Votes
    WHERE Postid = P.Id
      AND Votetypeid = 3)AS [Downvotes]
     FROM Posts Q
          INNER JOIN Posts P ON Q.Acceptedanswerid = P.Id
    WHERE P.Posttypeid = 2)A
 ORDER BY A.Score ASC;

Answer 2

为什么不将upvotes和downvotes视为一张桌子？然后，您可以按照自己的意愿工作。

您可以通过两种方式实现：表值函数或内联表。选择代码应该是这样的：

SELECT TOP 25 a.Id as [Post Link],
          a.Score as Score,
          [Upvotes].CountVotes as Upvotes,
          [Downvotes].CountVotes as DOwnvotes,
          (15 + (([Upvotes].CountVotes * 10) - ([Downvotes].CountVotes * 2))) AS    [Answerer's Reputation]
  FROM Posts q INNER JOIN Posts a ON q.AcceptedAnswerId = a.Id, 
    (SELECT COUNT(*) as CountVotes
           FROM Votes
           WHERE PostId = a.Id AND
                 VoteTypeId = 2) AS [Upvotes],

          (SELECT COUNT(*) as CountVotes
           FROM Votes
           WHERE PostId = a.Id AND
                 VoteTypeId = 3) AS [Downvotes]
WHERE a.PostTypeId = 2
ORDER BY a.Score ASC

使用所选值作为参数

2 个答案: