带有 COUNT 的 SQL 查询，具有 Count >1，显示重复项的完整详细信息

Question

我有一张这样的桌子：

name  employment_Status  email
----     ----             -----
David     E              David@email.com
John      U              John@email.com
Michael   E              Michael@email.com
Steve     E              Michael@email.com
James     U              David@email.com
Mary      U              Mary@email.com
Beth      E              Beth@email.com

我首先选择了email 和count(email)：

SELECT email, COUNT(email) AS emailCount
FROM Table
GROUP BY email
HAVING ( COUNT(email) > 1 );

当我尝试包含name 时出现问题：

SELECT name, email, COUNT(email) AS emailCount
FROM Table
GROUP BY name, email
HAVING ( COUNT(email) > 1 );

我想查找所有电子邮件地址重复的人，（仅在两个人都受雇的情况下 (E)）。但是它返回的结果为零。

我希望能够为具有重复电子邮件且拥有employment_Status E 的人显示所有信息。如果两个人有相同的电子邮件，但其中一个或两个都是失业者 (U)，则忽略。

谁能给点建议？

Answer 1

我想你想要exists:

select t.*
from t
where t.employeed = 'E' and
     exists (select 1
             from t t2
             where t2.email = t.email and t2.employeed = 'E' and
                   t2.name <> t.name
            );

请注意，这假定name（或至少name/email）是唯一的。

在 MySQL 8+ 中，您可以使用窗口函数：

select t.*
from (select t.*, count(*) over (partition by t.email) as cnt
      from t
      where t.employeed = 'E'
     ) t
where cnt >= 2;

Answer 2

一种方法是将您的查询用作 FROM 子句中的子查询，并将结果与​​主表连接。

SELECT t.*, d.emailCount
FROM (
    SELECT email, employment_Status, COUNT(*) AS emailCount
    FROM my_table
    GROUP BY email
    WHERE employment_Status = 'E'
    HAVING emailCount > 1 
) d
JOIN my_table t USING(email, employment_Status)

您也可以使用GROUP_CONCAT(name)，如果您可以在（逗号）分隔的字符串中获取名称：

SELECT email, COUNT(*) AS emailCount, GROUP_CONCAT(name) as names
FROM my_table
GROUP BY email
WHERE employment_Status = 'E'
HAVING emailCount > 1 

您的样本数据的结果是：

email               emailCount    names
-----------------------------------------------
Michael@email.com       2         Michael,Steve