MySQL - 选择用户不拥有的所有行

Question

我有一个小的MySQL问题。我想选择我的表中包含用户不拥有的数据的所有行（这是一个交换工具）。

实施例：

id | owner | event 
---------------------- 
1 | 4 | 3 
2 | 5 | 3 
3 | 3 | 2 
4 | 5 | 6 

我所有者＃4。我希望看到的，除了我已经拥有的那些所有的事件，所以我希望得到这样一个结果：

event | offered by 
------------------- 
2  |  3 
6  |  5 

是可以通过SQL选择这还是我要选择所有用户自己的事件，并删除其他所有与结果相同的事件id？

Answer 1

SELECT a.event, a.owner AS `Offered By` 
FROM tableName a 
WHERE a.event NOT IN 
     (
      SELECT b.event 
      FROM TableName b 
      WHERE owner = 4 
     ) 

或通过使用JOIN我更优选，

SELECT a.event, a.owner AS `Offered By` 
FROM tableName a 
     LEFT JOIN tableName b 
      ON a.event = b.event AND 
       b.owner = 4 
WHERE b.event IS NULL 

• SQLFiddle Demo (both queries)

输出

╔═══════╦════════════╗ 
║ EVENT ║ OFFERED BY ║ 
╠═══════╬════════════╣ 
║  2 ║   3 ║ 
║  6 ║   5 ║ 
╚═══════╩════════════╝ 

Answer 2

SELECT event, offered_by FROM events_table WHERE owner <> 4 

，或者如果它们是例如events 2表和offers

SELECT tb1.event, tb1.offered by FROM offers as tb1 
INNER JOIN events as tb2 on tb1.event = tb2.event 
WHERE tb2.owner <> 4 

Answer 3

我只是假设PHP这里

SELECT event 
FROM EXAMPLE_TABLE 
WHERE owner NOT IN '$owner_id'