【leetcode Database】196. Delete Duplicate Emails

題目：

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  | [email protected] |
| 2  | [email protected]  |
| 3  | [email protected] |
+----+------------------+
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  | [email protected] |
| 2  | [email protected]  |
+----+------------------+

解析：

最開始想到的做法是得到所有重複的Email地址的最小ID，然後把該Email地址的所有其它ID全部刪除，代碼如下：

# 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，卻報瞭如下的錯誤：You can't specify target table 'Person' for update in FROM clause

查詢之後得知，原因是不能對同一個表先select，之後再做update操作，需要加一箇中間表。修改後代碼如下：

# 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);

這次，可以accept了。但是，總覺得代碼太繁瑣。重新思考之後，其實我們可以把所有Email地址的最小ID全部都檢索出來，不管是否有重複。之後，再把其餘所有的ID都刪除就好了。修改後代碼如下：

# 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);