SQL 使用 COUNT 进行查询，计数 >1，显示重复项的完整详细信息

Question

我有一个 table 喜欢 :

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 email 和 count(email):

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

当我也尝试包含 name 时出现问题：

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

我想找到所有具有重复电子邮件地址的人（仅当两人都受雇时 (E)）。但是它返回零结果。

我希望能够显示具有重复电子邮件且 employment_Status E 的人的所有信息。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 (U)，则忽略即可。

有人可以指点一下吗？

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，这假设 name（或至少 name/email）是唯一的。

在MySQL 8+中，您可以使用window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 table.

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

您的示例数据的结果将是：

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

SQL 使用 COUNT 进行查询，计数 >1，显示重复项的完整详细信息

SQL Query with COUNT, Having Count >1, display full details of duplicates

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