a表是b表的子集，怎么得到b-a的记录，有这样的函数吗？求助！

[alex5006]

a表是b表的子集，怎么得到b-a的记录，有这样的函数吗？最好一句sql实现，不要用游标，那样效率太低了。有什么好的解决方法吗？谢谢！！！

[CrazyFor]

select b.* from b left join a on b.字段1=a.字段1 and b.字段2=a.字段2....
where a.任一字段 is null


b.字段1=a.字段1 and b.字段2=a.字段2.... --这是确认是子集的条件

[happydreamer]

差：
c1-c2:

select * from t1 where not exists(select 1 from t2 where t1.c1=t2.c1 and t1.c2=t2.c2)

c2-c1:

select * from t2 where not exists(select 1 from t1 where t1.c1=t2.c1 and t1.c2=t2.c2)

[pengdali]

select * from b where 主键 not in (select 主键 from a)

或：

差：
c1-c2:

select * from t1 where not exists(select 1 from t2 where t1.c1=t2.c1 and t1.c2=t2.c2)

c2-c1:

select * from t2 where not exists(select 1 from t1 where t1.c1=t2.c1 and t1.c2=t2.c2)

[hjb111]

是关系的差运算：
select * from b where not exists(select 1 from a where b.c1=a.c1 and b.c2=a.c2)

[txlicenhe]

select * from b where 主键 not in (select 主键 from a)