mysql - 从 SELECT 查询中得到相反的结果

Question

这些是我的表格:

`room`(roomID,roomNum)  
`customer`(customerID,Surname,etc)  
`contract`(contractID,roomID,weekNum)  
`paymen`t(paymentID,customerID,contractID,YearKoino)  

当我使用以下查询时:

`select` room.roomnum  
`from` payment,contract,room,customer  
`where` payment.contractID = contract.contractID  
`and` contract.roomID=room.roomID  
`and` customer.customerID=payment.customerID  
`and` contract.weeknum='40'  
`and` payment.YearKoino='2007' ;  

我得到的结果是:

+---------+  
| roomnum |  
+---------+  
| Δ-12    |  
| Γ-22    |  
| Α-32    |  
| Γ-21    |  
| Δ-11    |  
| Ε-12    |  
| Γ-31    |  
| Ε-22    |  
| Α-22    |  
| Δ-12    |  
| Γ-12    |  
+---------+  
11 rows in set  

我想做的是运行一个查询，给出完全相反的结果(表房间中的房间号不在表付款中)。这可以通过将上述查询的房间号结果与列 roomnum 进行比较来完成在房间的 table 上。到目前为止我的一些努力:

`Select` room.roomnum  
`from` room  
`where` NOT EXISTS  
(`select` room.roomnum  
`from` payment,contract,room,customer  
`where` payment.contractID = contract.contractID  
`and` contract.roomID=room.roomID  
`and` customer.customerID=payment.customerID  
`and` contract.weeknum='40'  
`AND` payment.YearKoino='2007');  
Empty set  

和

`SELECT` *
`FROM` customer a
`LEFT OUTER JOIN` payment b
`on` a.customerID=b.customerID
`where` a.customer is null;

我还尝试用“NOT IN”替换“NOT EXISTS”，但徒劳无功。我读到最好的方法是使用“left join”。好吧，当我有与简单的表进行比较。但在我的示例中，我必须将一列与表连接中的一列进行比较...

如有任何建议，我们将不胜感激。

Answer 1

我不确定为什么您的 not in 不起作用。

这应该有效(不使用表名别名):

   Select r1.roomnum 
   from room AS r1
   where r1.roomnum NOT IN 
       (select r2.roomnum 
        from payment,contract,room as r2,customer 
        where payment.contractID = contract.contractID
        and contract.roomID=r2.roomID 
        and customer.customerID=payment.customerID 
        and contract.weeknum='40' 
        AND payment.YearKoino='2007');