MySQL - 在WHERE子句中使用COUNT(*)？

Question

好的,我想找到所有具有相同值的行.(或者至少一对)

IE

James| 19.193.283.19
John| 20.134.232.344
Jack| 19.193.283.19 
Jonny| 19.193.283.19

我希望它返回James,Jack和Jonny的行 - 其中不止一行有IP '19 .193.283.19'.

我尝试做其他类似问题的答案:

select *
from `Zombie`
group by `Ip`
having count(*) > 1
order by `Ip` desc

但是它只返回了一排有一对或更多相似的'Ip'我想要的每一行.

我如何修改SQL,以便返回所有的indisinct行？

非常感谢.

Answer 1

您可以使用exists子查询来查找具有相同行的相同行Ip:

select  *
from    YourTable as yt1
where   exists
        (
        select  *
        from    YourTable as yt2
        where   yt1.name <> yt2.name
                and yt1.Ip = yt2.Ip
        )

使用相同的行数排序Ip可以使用自联接进行排序,例如:

select  yt1.name
,       yt1.Ip
from    YourTable as yt1
join    YourTable as yt2
on      yt1.name <> yt2.name
        and yt1.Ip = yt2.Ip
group by
        yt1.name
,       yt1.Ip
order by
        count(yt2.name) desc