1

我有一个这样的查询:

SELECT recipientid AS ID,
COUNT(*) AS Recieved FROM Inbox
GROUP BY recipientid

UNION

SELECT SenderId,
COUNT(*) AS [Sent] FROM Inbox
GROUP BY SenderId

输出:

RecipientID  Recieved

001             3
001             4
002             4
002             2
003            18
003            55

我如何重写是这样的,它显示如下:

RecipientID  Recieved  Sent

001             3       4
002             4       2
003            18       55

谢谢。

4

4 回答 4

3

只需加入子查询:

select a.ID,Received,Sent
from(
  SELECT recipientid AS ID,
  COUNT(*) AS Recieved FROM Inbox
  GROUP BY recipientid
)a
full outer join(
  SELECT SenderId as ID,
  COUNT(*) AS [Sent] FROM Inbox
  GROUP BY SenderId
)b
on (a.ID = b.ID)
order by a.ID;

请注意,这会获取任何收件人或发件人的所有sentreceived值。如果您只想要ID属于收件人和发件人的 s 的结果,请执行inner join.

于 2013-03-15T18:25:24.873 回答
2

我会source在您的查询中添加一列并做一个简单的数据透视

select ID, 
       max (case when source=1 then Cnt else 0 end) as Received,
       max (case when source=2 then Cnt else 0 end) as Sent
from (
  SELECT 1 as Source, 
         recipientid AS ID,
         COUNT(*) AS Cnt 
  FROM Inbox
  GROUP BY recipientid
  UNION
  SELECT 2 as Source, 
         SenderId,
         COUNT(*)  
  FROM Inbox
  GROUP BY SenderId
  ) x
GROUP BY ID
于 2013-03-15T18:27:19.647 回答
0

如果是 Postgres、MS SQL 或其他支持 CTE -

With Both as
(
SELECT
  recipientid AS ID,
  Count(*) AS Recieved,
  0 as [Sent] 
FROM Inbox
GROUP BY recipientid
UNION
SELECT
  SenderId as ID,
  0 as Recieved,
  Count(*) AS [Sent]
FROM Inbox
GROUP BY SenderId
)
SELECT
  ID,
  Sum(Received) as [Received],
  Sum(Sent) as [Sent]
FROM BOTH
GROUP BY ID
ORDER BY 1
于 2013-03-15T18:26:25.507 回答
0

假设您有一个users带有 ID 的表,您可以执行以下操作:

SELECT
    users.id,
    COUNT(sent.senderid) AS sent,
    COUNT(received.recipientid) AS received
FROM
    users
    LEFT JOIN inbox AS sent ON sent.senderid = users.id
    LEFT JOIN inbox AS received ON received.recipientid = users.id
GROUP BY sent.senderid, received.recipientid
ORDER BY users.id;
于 2013-03-15T18:44:11.473 回答