MySQL的：不存在匹配的記錄

Question

我有兩個表：

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

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

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

如何從客戶表中選擇行，其中Customer.VoicemailID不再是Vo icemail表中的有效記錄？

這是一種情況，即記錄曾經存在於Voicemail表中，但此後已被刪除。我現在需要找到Customer表中有VoicemailID不存在記錄的所有記錄。

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結果我想，與其中的語音郵件的實例仍然存在導致混合。

Answer 1

這應該給你你想要的結果：

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

Answer 2

您正處於正確的軌道上，LEFT JOIN。但是你需要尋找匹配，然後在失敗時返回：

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

Answer 3

您可以使用子查詢並查找不在那裏的行。檢查MySQL的documentation爲WHERE NOT EXISTS語法：

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

您可以使用相關子查詢這樣

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

Answer 6

這裏是[PHP：5.5.3]中的正確查詢，我在Xampp 1.8.3中使用phpMyAdmin。

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