mysql - 匹配来自两个不同表的值

Question

给定 2 个表，TableA 和 TableB，当“值”与 TableB 的任何“值”都不匹配时，如何获取 TableA 的“id”？

表 A

+----+-------+
| id | value |
+----+-------+
| 1  |   a   |
| 2  |   b   |
| 3  |   c   |
| 4  |   d   |
| 5  |   e   |
+----+-------+

表 B

+----+-------+
| id | value |
+----+-------+
| 1  |   a   |
| 2  |   b   |
| 3  |   c   |
| 4  |   c   |
| 5  |   f   |
+----+-------+

结果表

+----+
| id |
+----+
| 4  |
| 5  |
+----+

已编辑（SQLFiddle）： http ://sqlfiddle.com/#!2/4c8c9

score 5 · Accepted Answer

如果要验证两个表中的 id/值是否相同，请使用：

select distinct id
from tablea
where (id, value) not in (select id, value
                          from tableb)

如果您只想比较值：

select distinct id
from tablea
where (value) not in (select value
                          from tableb)

score 3 · Accepted Answer

select a.id
from tableA a
left outer join tableB b on a.id = b.id and a.value = b.value
where a.id is not null and b.value is null

2 回答 2