We read every piece of feedback, and take your input very seriously.
To see all available qualifiers, see our documentation.
Have a question about this project? # for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “#”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? # to your account
Write a SQL query to find all duplicate emails in a table named Person.
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.
这道题让我们求重复的邮箱,那么最直接的方法就是用Group by...Having Count(*)...的固定搭配来做,代码如下:
解法一:
SELECT Email FROM Person GROUP BY Email HAVING COUNT(*) > 1;
我们还可以用内交来做,用Email来内交两个表,然后返回Id不同的行,则说明两个不同的人使用了相同的邮箱:
解法二:
SELECT DISTINCT p1.Email FROM Person p1 JOIN Person p2 ON p1.Email = p2.Email WHERE p1.Id <> p2.Id;
参考资料:
https://leetcode.com/discuss/53206/a-solution-using-a-group-by-and-another-one-using-a-self-join
LeetCode All in One 题目讲解汇总(持续更新中...)
The text was updated successfully, but these errors were encountered:
No branches or pull requests
Write a SQL query to find all duplicate emails in a table named
Person
.For example, your query should return the following for the above table:
Note : All emails are in lowercase.
这道题让我们求重复的邮箱,那么最直接的方法就是用Group by...Having Count(*)...的固定搭配来做,代码如下:
解法一:
我们还可以用内交来做,用Email来内交两个表,然后返回Id不同的行,则说明两个不同的人使用了相同的邮箱:
解法二:
参考资料:
https://leetcode.com/discuss/53206/a-solution-using-a-group-by-and-another-one-using-a-self-join
LeetCode All in One 题目讲解汇总(持续更新中...)
The text was updated successfully, but these errors were encountered: