目錄

一、196. 刪除重復的電子郵箱 - 力扣（LeetCode）

二、602. 好友申請 II ：誰有最多的好友 - 力扣（LeetCode）

三、176. 第二高的薪水 - 力扣（LeetCode）

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一、196. 刪除重復的電子郵箱 - 力扣（LeetCode）

題意就是刪除刪除重復的郵箱

很容易可以想到 delete from person where id in (一坨)

繞了個彎子 讓你寫刪除語句本質還是寫查詢語句

1. 第一層查詢使用窗口函數 分組加排序

   select *,
   row_number() over(partition by email order by id asc) as 'rank' 
   from Person

   

2. 可以顯然得出 臨時表中rank >1 的都是重復的，嵌套一層查id出來

    select id from (select *,row_number() over(partition by email order by id asc) as 'rank' from Person) temp  where temp.rank = 1

3. 執行刪除語句

   delete from Person where id not in (select id from (select *,row_number() over(partition by email order by id asc) as 'rank' from Person) temp  where temp.rank = 1
   )

   

二、602. 好友申請 II ：誰有最多的好友 - 力扣（LeetCode）

?

with t1 as(select requester_id as 'id' from RequestAcceptedunion allselect accepter_id  as 'id' from RequestAccepted
),
t2 as(select id,count(id) over(partition by id rows between unbounded preceding and unbounded following) as 'num'from t1
),
t3 as(select *,dense_rank() over(partition by null order by num desc) as 'rank'from t2
)
select id,num 
from t3 
where t3.rank = 1
limit 1

理解就是加好友是相互的！！！！！！！！

?把兩列數據并成一列? 然后窗口函數分組排序

三、176. 第二高的薪水 - 力扣（LeetCode）

也是窗口函數分組排序? 但是這個題就比較麻煩 需要考慮空結果集輸出null

select ifnull((with t1 as(select *,dense_rank() over(partition by null order by salary desc) as 'rank'from Employee),t2 as(select distinct salary as 'SecondHighestSalary' from t1where t1.rank = 2)select SecondHighestSalary from t2
),null) as 'SecondHighestSalary'

結束三道sql！