感谢 2022612564 提供了答案。--直接找出重复的
select a.* from `guest` a join
(SELECT id_k ,ip FROM guest where id_k = N GROUP BY id_k ,ip HAVING count(ip) > 1)b
on a.id_k=b.id_k and a.ip=b.ip 其实也可以反过来去除出单个的,剩下的必然是重复的
select * from guest where ip not in(select ip from guest where id_k=N group by ip having count(ip)=1) and id_k=N
select a.* from `guest` a join
(SELECT id_k ,ip FROM guest where id_k = N GROUP BY id_k ,ip HAVING count(ip) > 1)b
on a.id_k=b.id_k and a.ip=b.ip 其实也可以反过来去除出单个的,剩下的必然是重复的
select * from guest where ip not in(select ip from guest where id_k=N group by ip having count(ip)=1) and id_k=N
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