技術標籤：SQLmysqlgroup byhaving

一、tim題目描述

編寫一個 SQL 查詢，查詢 Person 表中所有重複的電子郵箱。
示例：
±—±--------+
| Id | Email |
±—±--------+
| 1 | [email protected] |
| 2 | [email protected] |
| 3 | [email protected] |
±—±--------+
根據以上輸入，你的查詢應返回以下結果：
±--------+

| Email |
±--------+
| [email protected] |
±--------+

題解：
方法一：group by + having
優先順序：where>group by>having>order by

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

方法二：group by + 子查詢

SELECT 
    t.Email 
FROM 
    (SELECT Email,count(*) sumEmail
    FROM
 
 Person
    GROUP BY Email) t
WHERE t.sumEmail>1;