SQL

Question

我正在尝试 select table 中具有相同 Email 但具有不同 Name 的所有用户。到目前为止，我已经设法获取了所有重复的行 Email，但我仍坚持下一步。

SELECT * FROM users WHERE Email IN
(SELECT Email FROM users GROUP BY Email HAVING COUNT(*) > 1)

提前致谢

Answer 1

您可以尝试以下方法：

SELECT u.* 
FROM users u
LEFT JOIN (
   SELECT Email  
   FROM users
   GROUP BY Email 
   HAVING COUNT(*) = COUNT(DISTINCT name)
) tmp ON u.Email = tmp.Email
WHERE tmp.Email IS NOT NULL

Answer 2

试试这个

SELECT * 
FROM users U1
INNER JOIN 
users U2
on U1.Email=U2.Email
AND U1.Name <> U2.Name

Answer 3

我想你只想在子查询中使用 count(distinct name):

SELECT *
FROM users
WHERE Email IN (SELECT Email
                FROM users
                GROUP BY Email
                HAVING COUNT(distinct Name) > 1
               ) ;

对于 having 子句，我更喜欢 having min(name) <> max(name)。效率稍微高一点。

然而，最有效的方法可能是使用 window 函数：

select u.*
from (select u.*, min(name) over (partition by email) as minname,
             max(name) over partition by email) as maxname
      from users u
     ) u
where minname <> maxname;

SQL - Select 所有具有相同电子邮件但不同名称的行

SQL - Select all rows that have the same Email but different Name

select

count

group-by

duplicates