3 计数函数返回相同的值答案

【问题标题】：3 Count functions returning the same value3 计数函数返回相同的值
【发布时间】：2018-08-07 02:15:35
【问题描述】：

所以我有这个查询：

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];

应该返回结果：

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

但我得到的是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，但是我如何更改它以获得我想要的结果？

【问题讨论】：

ms访问还是mysql????
语法明显是MS Access，所以我调整了标签。

标签： sql database ms-access count

【解决方案1】：

您误解了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)。因为 MS-Access 不支持COUNT(DISTINCT)，您可能会将其解释为“获取功能数据库”。

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];

【讨论】：

【解决方案2】：

当您加入时，每个组合都会获得一行。因此，count 确实计算每一行，其值不同于 null。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];

如果系统不支持这种语法，你可能需要做子查询：

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;

【讨论】：