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目录

一、196. 删除重复的电子邮箱 - 力扣（LeetCode）

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

三、176. 第二高的薪水 - 力扣（LeetCode）

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一、196. 删除重复的电子邮箱 - 力扣（LeetCode）

题意就是删除删除重复的邮箱

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

绕了个弯子 让你写删除语句本质还是写查询语句

1. 第一层查询使用窗口函数 分组加排序

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

   

2. 可以显然得出 临时表中rank >1 的都是重复的，嵌套一层查id出来

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

3. 执行删除语句

   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. 好友申请 II ：谁有最多的好友 - 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！！！！！！！！

 把两列数据并成一列  然后窗口函数分组排序

三、176. 第二高的薪水 - 力扣（LeetCode）

也是窗口函数分组排序  但是这个题就比较麻烦 需要考虑空结果集输出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'

结束三道sql！