[LeetCode] 182.Duplicate Emails 重復的郵箱
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
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