Question

所以我有这个问题：

SELECT Posts.[Post ID], Count(Comments.[Comment ID]) AS [CountOfComment ID],
Count(Shares.[Share ID]) AS [CountOfShare ID],
Count(Likes.[Like ID]) AS [CountOfLike ID]

FROM ((Posts INNER JOIN Likes ON Posts.[Post ID] = Likes.[Post ID]) 
INNER JOIN Shares ON Posts.[Post ID] = Shares.[Post ID]) 
INNER JOIN Comments ON Posts.[Post ID] = Comments.[Post ID]

WHERE posts.[post id] = 5 

GROUP BY Posts.[Post ID];

哪个应该返回结果：

post.[post id]= 5 
CountofcommentID= 4
CountofShareID = 5
CountofLikeID = 15

但我获得post.[post id] = 5, and the rest =300.

我知道这是因为它正在计算所有ID的所有排列，但我如何更改它以获得我想要的结果？

Answer 1

加入时，每个组合都会有一行。因此，count会计算每一行，其值与null不同。所以，你想要的是计算DISTINCT ID。即：

SELECT Posts.[Post ID], Count(DISTINCT Comments.[Comment ID]) AS [CountOfComment ID],
Count(DISTINCT Shares.[Share ID]) AS [CountOfShare ID],
Count(DISTINCT Likes.[Like ID]) AS [CountOfLike ID]

FROM ((Posts INNER JOIN Likes ON Posts.[Post ID] = Likes.[Post ID]) 
INNER JOIN Shares ON Posts.[Post ID] = Shares.[Post ID]) 
INNER JOIN Comments ON Posts.[Post ID] = Comments.[Post ID]

WHERE posts.[post id] = 5 

GROUP BY Posts.[Post ID];

如果系统不支持此语法，则可能需要执行子查询：

Select Posts.[Post ID], 
   (Select count([Share ID]) from Shares where Posts.[Post ID] = Shares.[Post ID]) as SharesCount
FROM Posts

WHERE posts.[post id] = 5;

Answer 2

你误解了COUNT()的作用。它计算非NULL值的数量。正确的解决方案是使用COUNT(DISTINCT)。由于MS-Access不支持COUNT(DISTINCT)，您可能会将其解释为＆＃34;获取功能数据库＆＃34;。

在MS Access中，您可以使用某种类型的子查询来执行此操作。这是一种方法：

SELECT p.[Post ID],
       (SELECT COUNT(*)
        FROM (SELECT DISTINCT [Post Id], [Comment ID]
              FROM Comments as c
             ) as c
        WHERE c.[Post Id] = p.[Post Id]
       ) as [CountOfComment ID],
       (SELECT COUNT(*)
        FROM (SELECT DISTINCT [Post Id], [Share ID]
              FROM Shares as s
             ) as s
        WHERE s.[Post Id] = p.[Post Id]
       ) as [CountOfShare ID],
       (SELECT COUNT(*)
        FROM (SELECT DISTINCT [Post Id], [Like ID]
              FROM Likes as l
             ) as l
        WHERE l.[Post Id] = p.[Post Id]
       ) as [CountOfLikes ID]
FROM Posts as p
WHERE p.[post id] = 5 
GROUP BY p.[Post ID];

3计数函数返回相同的值

2 个答案: