leetcode 182

发布时间 2023-04-11 15:46:57作者: Carl_ZhangJH

查找重复的电子邮箱

 

select email as Email from Person group by email having count(email) > 1

 

select email as Email from (
    select email ,count(email) as c from Person group by email
) r where r.c > 1

 

select distinct p1.email as Email from Person p1
left join Person p2
on p1.email = p2.email
where p1.id != p2.id

 

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