[LeetCode] 182.Duplicate Emails 重復(fù)的郵箱

Write a SQL query to find all duplicate emails in a table named Person.

+----+---------+
| Id | Email   |
+----+---------+
| 1  | a@b.com |
| 2  | c@d.com |
| 3  | a@b.com |
+----+---------+

For example, your query should return the following for the above table:

+---------+
| Email   |
+---------+
| a@b.com |
+---------+

Note: All emails are in lowercase.

這道題讓我們求重復(fù)的郵箱，那么最直接的方法就是用Group by...Having Count(*)...的固定搭配來做，代碼如下：

解法一：

SELECT Email FROM Person GROUP BY Email
HAVING COUNT(*) > 1;

我們還可以用內(nèi)交來做，用Email來內(nèi)交兩個表，然后返回Id不同的行，則說明兩個不同的人使用了相同的郵箱：

解法二：

SELECT DISTINCT p1.Email FROM Person p1
JOIN Person p2 ON p1.Email = p2.Email
WHERE p1.Id > p2.Id;

參考資料：

https://leetcode.com/discuss/53206/a-solution-using-a-group-by-and-another-one-using-a-self-join

到此這篇關(guān)于SQL實現(xiàn)LeetCode(182.重復(fù)的郵箱)的文章就介紹到這了,更多相關(guān)SQL實現(xiàn)重復(fù)的郵箱內(nèi)容請搜索腳本之家以前的文章或繼續(xù)瀏覽下面的相關(guān)文章希望大家以后多多支持腳本之家！