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”不匹配时，如何获得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

Answer 1

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)

请参阅SQL Fiddle with Demo

如果您只想比较值：

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

请参阅SQL Fiddle with Demo

Answer 2

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

SQLFiddle demo