202553-sql - 技术栈

[一、196. 删除重复的电子邮箱 - 力扣（LeetCode）](#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）](#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）](#三、176. 第二高的薪水 - 力扣（LeetCode）)

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
第一层查询使用窗口函数分组加排序
sql 复制代码
select *,
row_number() over(partition by email order by id asc) as 'rank' 
from Person
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
sql 复制代码
 select id from (
        select *,
                row_number() over(partition by email order by id asc) as 'rank' from Person
    ) temp  where temp.rank = 1
执行删除语句
sql 复制代码
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）

sql 复制代码

with t1 as(
    select requester_id as 'id' from RequestAccepted
    union all
    select 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

sql 复制代码

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 t1
        where t1.rank = 2
    )
    select SecondHighestSalary from t2
),null) as 'SecondHighestSalary'

结束三道sql！