我有这四张桌子。
Mail
----
id
date
Source (can have many source)
------
id
mailID : FK Mail.id
personID : FK Person.id
Destination (can have many destination)
-----------
id
mailID : FK Mail.id
personID : FK Person.id
Person
------
id
name 我正在尝试创建一个查询,它返回所有源ids和目标ids的每个邮件。
到目前为止,我已经提出了这个查询,但是由于它两次查询同一个表,所以效率很低。
Select * FROM (
Select m.*, personID, 'Source' AS TableName
FROM Mail m join Source s
ON m.id = s.mailid
UNION ALL
Select m.*, personID, 'Destination' AS TableName
FROM Mail m join Destination d
ON m.id = d.mailid
) t ORDER BY id, TableName;
results:
id | date | personID | TableName
--------------------------------
1 | 1-1-11| 3 | Source
1 | 1-1-11| 4 | Source
1 | 1-1-11| 5 | Source
1 | 1-1-11| 10 | Destination
1 | 1-1-11| 11 | Destination
2 | 2-2-11| 1 | Source
2 | 2-2-11| 2 | Destination发布于 2015-12-18 10:11:07
试试这个:
SELECT m.id, m.name, a.personID, a.TableName
FROM Mail m
INNER JOIN (SELECT s.mailid, s.personID, 'Source' AS TableName
FROM Source s
UNION ALL
SELECT d.mailid, d.personID, 'Destination' AS TableName
FROM Destination d
) AS a ON m.id = a.mailid
ORDER BY m.id, a.TableName;https://stackoverflow.com/questions/34353023
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