比较日期和ID？

Question

使用SQL Server 2000

我想获得Table2.TimeIn Table2.TimeOut根据Table1.personid和也如果Table1.Date = Table3.Date那么它应该需要一个Table3.TimeIn，Table3.TimeOut。

3表

表1

ID Date 

001 20090503 
001 20090504 
001 20090506 
002 20090505 
002 20090506 

...等，

表2

ID TimeIn TimeOut 

001 08:00:00 18:00:00 
002 08:00:00 21:00:00 

...等，

表3

ID Date TimeIn TimeOut 

001 20090504 10:00:00 22:00:00 
001 20090505 17:00:00 23:00:00 
002 20090505 12:00:00 21:00:00 

...等，

Select Table1.ID, 
     Table1.Date, 
     Table2.TimeIn, 
     Table2.TimeOut 
    from Table1 
Inner Join Table2 on Table1.ID = Table2.ID 

如果Table1.Date = Table3.Date那么就应该采取Table3.TimeIn，Table3.TimeOut其他Table2.TimeIn，Table2.Timeout

的预期输出

ID Date TimeIn TimeOut 

001 20090503 08:00:00 18:00:00 
001 20090504 10:00:00 22:00:00 
001 20090506 08:00:00 18:00:00 
002 20090505 12:00:00 21:00:00 
002 20090506 08:00:00 21:00:00 

...等，

如何为这种情况编写查询？

Answer 1

员工时间表后备?:

SELECT Table1.ID 
    ,Table1.Date 
    ,COALESCE(Table3.TimeIn, Table2.TimeIn) AS TimeIn 
    ,COALESCE(Table3.TimeOut, Table2.TimeOut) AS TimeOut 
FROM Table1 
INNER JOIN Table2 -- Always have an expected schedule for an employee 
    ON Table1.ID = Table2.ID 
LEFT JOIN Table3 -- May.may not have an actual schedule for an employee 
    ON Table3.ID = Table1.ID 
    AND Table3.Date = Table1.Date 
/* 
ORDER BY Table1.ID 
    ,Table1.Date 
*/