如何使用自我联接获取不匹配的记录

Question

假设我有一个名为users的表，其中包含以下各列：

 user_id (int), 
 job_id (int),
 created (date)

我想抓住两个用户，并根据job_id获得不匹配的记录。

例

user_id  | job_id | created 

15242    |  234   | 2015-04-07 
15242    |  441   | 2015-04-08
15242    |  345   | 2015-04-08
24521    |  234   | 2015-04-09

我想获得job_ids 441和345。

因此需要自我加入

SELECT users.job_id
FROM users as switch_from
LEFT JOIN users as switch_to ON switch_from .job_id = switch_to .job_id
WHERE switch_from.user_id = 15242 AND switch_to.user_id = 24521;

如果不是这给我的别名表中的所有job_ids switch_from从别名表中缺少switch_to ？

这将返回唯一具有匹配job_id的行。

Answer 1

我认为您也可以使用减号语句：

SELECT member_id, name FROM a
MINUS
SELECT member_id, name FROM b

如果您希望switch_to中的记录在switch_from中是nog，则可以这样进行。

SELECT users.job_id FROM users as switch_to
WHERE switch_to.user_id = 24521
MINUS
SELECT users.job_id FROM users as switch_from
WHERE switch_from.user_id = 15242;

Answer 2

像这样尝试：

SELECT switch_from.job_id
FROM users as switch_from
LEFT JOIN users as switch_to ON switch_from.job_id = switch_to.job_id AND switch_to.user_id = 24521
WHERE switch_from.user_id = 15242 AND switch_to.user_id IS NULL;

Answer 3

试试这个查询：

SELECT users.job_id FROM users as switch_from
   LEFT JOIN users as switch_to ON 
   ( switch_from .job_id = switch_to .job_id and switch_to.user_id = 24521 )
   WHERE switch_from.user_id = 15242