MySQL的：不存在匹配的记录

Question

我有两个表：

Customer 
+---+-----------+ 
|ID |VoicemailID| 

Voicemail 
+---+----------+ 
|ID |CustomerID| 

Voicemail.CustomerID releates到Customer.ID，反之亦然。

如何从客户表中选择行，其中Customer.VoicemailID不再是Vo icemail表中的有效记录？

这是一种情况，即记录曾经存在于Voicemail表中，但此后已被删除。我现在需要找到Customer表中有VoicemailID不存在记录的所有记录。

我想：

SELECT DISTINCT Customer.ID, Customer.VoicemailID 
FROM Customers LEFT JOIN Voicemail ON Customer.VoicemailID <> Voicemail.ID 

但是我相信它返回的结果我想，与其中的语音邮件的实例仍然存在导致混合。

Answer 1

这应该给你你想要的结果：

SELECT Customer.ID, Customer.VoicemailID 
FROM Customer 
    LEFT JOIN Voicemail ON Voicemail.ID = Customer.VoicemailID 
WHERE Voicemail.ID IS NULL 

Answer 2

您正处于正确的轨道上，LEFT JOIN。但是你需要寻找匹配，然后在失败时返回：

SELECT c.ID, c.VoicemailID 
FROM Customer c LEFT JOIN 
    Voicemail v 
    ON c.VoicemailID = v.ID 
WHERE v.ID IS NULL; 

Answer 3

您可以使用子查询并查找不在那里的行。检查MySQL的documentation为WHERE NOT EXISTS语法：

SELECT * 
FROM table 
WHERE NOT EXISTS(SELECT 1 
       FROM otherTable 
       WHERE table.id = otherTable.someField) 

Answer 4

select VoicemailID 
    from Customer 
    where VoicemailID not in (select id from Voicemail) 

Answer 5

您可以使用相关子查询这样

select * from customer a 
where not exists (select 1 from voicemail b 
        where b.id = a.voicemailid) 

Answer 6

这里是[PHP：5.5.3]中的正确查询，我在Xampp 1.8.3中使用phpMyAdmin。

SELECT * FROM Customer 
WHERE Customer.VoicemailID NOT IN (SELECT ID FROM Voicemail)