表:yd
字段:
bgid pay
1 1
1 1
1 0
2 1
3 0
要求: bgid 值相同时,pay全部为1 才输出
以上输出结果: bgid 2
回复讨论(解决方案)
有人会不,有人会不,
select * from yd where bgid not in (select bgid from yd where pay !=1)
select bgid, pay
from (select *, count(*) as c, sum(pay) as s from yd group by bgid) t
where c=s
select bgid,pay from yd where bgid not in(select bgid from yd where pay=0);
select count(select bgid from yd where pay=1 group by bgid) as result
不知有没有问题？
改正如下：
select bgid,count(*) as result from table_1 where pay=1 group by bgid
统计的是pay=1对应的bgid字段重复出现的次数