【leetcode Database】196. Delete Duplicate Emails

╰+哭是因爲堅強的太久メ 2022-09-25 01:16 97阅读 0赞

题目：

Write a SQL query to delete all duplicate email entries in a table named `Person`, keeping only unique emails based on its smallest Id.

+----+------------------+
    | Id | Email            |
    +----+------------------+
    | 1  | john@example.com |
    | 2  | bob@example.com  |
    | 3  | john@example.com |
    +----+------------------+
    Id is the primary key column for this table.

For example, after running your query, the above `Person` table should have the following rows:

+----+------------------+
    | Id | Email            |
    +----+------------------+
    | 1  | john@example.com |
    | 2  | bob@example.com  |
    +----+------------------+

解析：

最开始想到的做法是得到所有重复的Email地址的最小ID，然后把该Email地址的所有其它ID全部删除，代码如下：

# Write your MySQL query statement below
    DELETE FROM Person WHERE 
    Email IN (SELECT  Email FROM Person GROUP BY Email HAVING COUNT(Email) >1) 
    AND Id NOT IN (SELECT MIN(Id) FROM Person GROUP BY Email HAVING COUNT(Email) >1);

但是，却报了如下的错误： **You can't specify target table 'Person' for update in FROM clause**

查询之后得知，原因是不能对同一个表先select，之后再做update操作，需要加一个中间表。修改后代码如下：

# Write your MySQL query statement below
    DELETE FROM Person WHERE Email IN 
    (SELECT t.Email FROM (SELECT  Email FROM Person GROUP BY Email HAVING COUNT(Email) >1) t) 
    AND Id NOT IN (SELECT s.Id FROM (SELECT MIN(Id) AS Id FROM Person GROUP BY Email HAVING COUNT(Email) >1) s);

这次，可以accept了。但是，总觉得代码太繁琐。重新思考之后，其实我们可以把所有Email地址的最小ID全部都检索出来，不管是否有重复。之后，再把其余所有的ID都删除就好了。修改后代码如下：

# Write your MySQL query statement below
    DELETE FROM Person WHERE Id NOT IN (SELECT s.Id FROM (SELECT MIN(Id) AS Id FROM Person GROUP BY Email) s);