简而言之,我有一个看起来像这样的表:
ID - NAME - EMAIL
1 - JIM - Jim@hotmail.com
2 - JIM - Jim@gmail.com
3 - BARRY - Barry@hotmail.com
4 - JACK - Jack@hotmail.com
我想查询该表并仅返回NAME具有多个EMAIL值的字段,因此要求的结果是:
NAME - EMAIL
JIM - Jim@hotmail.com
JIM - Jim@gmail.com
其他两个(Barry和Jack)不需要返回,因为它们只有一个电子邮件值。
我在Access中的派生表设置方面有些挣扎,以下内容将不起作用,因为它在FROM子句中给我一个错误。还有其他语法吗?
select DISTINCT x.ID, x.NAME, x.EMAIL
from t as x
join (
select ID
from t
group by ID
having count(distinct EMAIL) > 1
) as y
on x.ID = y.ID
答案 0 :(得分:1)
尝试一下
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答案 1 :(得分:1)
尝试一下:
select x.ID, x.NAME, x.EMAIL
from t as x
where x.NAME in (
select t.NAME
from t
group by t.NAME
having count(*) > 1)