mysql - 在 MySQL 中获取不存在于另一个表的两列中的行

Question

MySQL 中有两个表，详细信息如下:

表 1:

+-------------+---------------+
 | user_id     |  isactive     |
 +-------------+---------------+
 | aaa         | 0             |
 +-------------+---------------+
 | bbb         | 0             | 
 +-------------+---------------+ 

表 2:

 +-------------+---------------+-----------+
 |store_no     |  owner        | store     |
 +-------------+---------------+-----------+
 |1234         | aaa           | aaa,xyz   |
 +-------------+---------------+-----------+
 |1006         | aaa           | aaa       | 
 +-------------+---------------+-----------+
 |1005         | ccc           | www       | 
 +-------------+---------------+-----------+

我需要从表 1 中获取条目既不在表 2 的“owner”列中也不在“store”列中的行。例如，在上面的场景中，结果集应该包含“bbb”。我尝试使用 find_in_set、locate 等，但无法按要求获取详细信息。请帮助..

更新表格格式

查询:

select a.user_id from table1 u
left outer join table2 a
on (owner=user_id or concat(',',store,',') like concat('%,',user_id,',%'))
where (find_in_set(user_id,owner) = 0 or find_in_set(user_id,store) = 0)
and  isactive=0

仅供引用，商店列可以具有多个用户 ID 的串联值

Answer 1

您可以尝试使用NOT EXISTS

SELECT 
T1.user_id
FROM TABLE_1 T1
WHERE NOT EXISTS (
    SELECT 1 
      FROM 
    TABLE_2 T2
    WHERE T2.owner = T1.user_id OR FIND_IN_SET(T1.user_id,T2.store) > 0
);

SQL FIDDLE DEMO

建议:

Is storing a delimited list in a database column really that bad?