select
a.name
,
count(price)
from
a,b
where
a.mediid
=
b.mediid
and
a.forgid
=
b.forgid
and
a.forgid
=
'E'
group
by
a.name
如果要去掉重复价格可以增加distinct,如下:
select
a.name
,
count(distinct
price)
from
a,b
where
a.mediid
=
b.mediid
and
a.forgid
=
b.forgid
and
a.forgid
=
'E'
group
by
a.name
问题2:查询E机构下面,两个表中编号相同的项目
select
count(name)
from
a,b
where
a.mediid
=
b.mediid
and
a.forgid
=
b.forgid
and
a.forgid
=
'E'
select a.d,b.b1 from A a left join B b on a.A1=b.B1 where a.C = ?(默认提供左连接写法,根据你的需求自己选择适合的连接)欢迎分享,转载请注明来源:内存溢出
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