1. 176. 第二高的薪水
筛选出第二大
查询并返回 Employee 表中第二高的薪水 。如果不存在第二高的薪水，查询应该返回 null(Pandas 则返回 None) 。查询结果如下例所示。
666中等的第一题就上强度
强行解法
select max(salary) as SecondHighestSalary from Employee
where salary!=(select max(salary) as salary from Employee);万一是其他次序的解法，使用limit（看2.）
1.标量子查询，查询为空的时候会填充null
2.
limit n,m等价于limit m offset n，表示跳过开头的n行，返回接下来的m条数据。降序之后的第2条数据就是limit 1,1
SELECT( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) AS SecondHighestSalary第三题提供的思路，但是无法实现重复的最大值
-- 每个部门第一第三高
SELECT S.NAME, S.EMPLOYEE, S.SALARYFROM (SELECT D.NAME,T.NAME EMPLOYEE,T.SALARY,ROW_NUMBER() OVER(PARTITION BY T.DEPARTMENTID ORDER BY T.SALARY DESC) RNFROM EMPLOYEE TLEFT JOIN DEPARTMENT DON T.DEPARTMENTID = D.ID) SWHERE S.RN = 1 OR S.RN = 32. 177. 第N高的薪水
写函数
查询 Employee 表中第 n 高的工资。如果没有第 n 个最高工资，查询结果应该为 null 。
MySQL高级-自定义函数-csdn
有点玄学的函数
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGINset n=n-1;RETURN (SELECT distinct salary FROM Employee ORDER BY salary DESC LIMIT N, 1);
END3. 178. 分数排名
考察同3.
select score, dense_rank() over(order by score desc) as 'rank' from Scores4. 180. 连续出现的数字
连续值
SELECT DISTINCT Num AS ConsecutiveNums FROM Logs
WHERE (Id+1, Num) IN (SELECT * FROM Logs)
AND (Id+2, Num) IN (SELECT * FROM Logs)5. 184. 部门工资最高的员工
最大值重复
select d.name as Department, e.name as Employee, e.salary as Salary
from Employee e left join Department d on e.departmentId = d.id
where (e.departmentId, e.salary) in (select distinct departmentId, max(salary) from Employee
group by departmentId)开窗函数实现
select Department, Employee, Salary from
(select d.name Department, e.name Employee, e.salary Salary, rank() over(partition by e.departmentId order by e.salary DESC) rankA
from Employee e left join Department d on e.departmentId = d.id) ranktbl
where rankA = 1SQL开窗函数（窗口函数）详解-CSDN博客
over() 一类是聚合开窗函数，一类是排序开窗函数。
调用格式为：函数名(列名) OVER(partition by 列名 order by列名) 。
聚合函数对一组值执行计算并返回单一的值，如sum()，count()，max()，min()， avg()等
常与group by子句连用。除了 COUNT 以外，聚合函数忽略空值
想知道各个地区的前几名、各个班的前几名。这时候需要每一组返回多个值
SQL 标准允许将所有聚合函数用作开窗函数，用OVER 关键字区分开窗函数和聚合函数
开窗函数与聚合函数一样，也是对行集组进行聚合计算
row_number () over()
ROW_NUMBER() OVER(PARTITION BY T.DEPARTMENTID ORDER BY T.SALARY DESC) RN但是这个无法取出重复的最大值
对相等的值不进行区分，其实就是行号，相等的值对应的排名不同，序号从1到n连续。
rank() over()：
相等的值排名相同，但若有相等的值，则序号从1到n不连续。如果有两个人都排在第3名，则没有第4名。1233567
dense_rank() over()：
对相等的值排名相同，但序号从1到n连续。如果有两个人都排在第一名，1123456
可以看作是把有序的数据集合平均分配到指定的数量n的桶中,将桶号分配给每一行，排序对应的数字为桶号，序号从1到n连续。如果不能平均分配，则较小桶号的桶分配额外的行，并且各个桶中能放的数据条数最多相差1。1111222333444555（放三轮，1桶多一个）
6. 550. 游戏玩法分析 IV
日期计算，查询相互运算
示例 1：输入：
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1 | 2 | 2016-03-01 | 5 |
| 1 | 2 | 2016-03-02 | 6 |
| 2 | 3 | 2017-06-25 | 1 |
| 3 | 1 | 2016-03-02 | 0 |
| 3 | 4 | 2018-07-03 | 5 |
+-----------+-----------+------------+--------------+
输出：
+-----------+
| fraction |
+-----------+
| 0.33 |
+-----------+
解释：
只有 ID 为 1 的玩家在第一天登录后才重新登录，所以答案是 1/3 = 0.33select round((select count(player_id) from Activity -- 算出所有条件玩家数where (player_id,event_date-INTERVAL 1 day) in (select player_id,min(event_date) from Activity group by player_id))/(select count(distinct player_id) from Activity) -- 算出所有玩家数,2) as fraction 7. 570. 至少有5名直接下属的经理
混进来一个简单题
# Write your MySQL query statement below
select e1.name from Employee e1 left join Employee e2 on e1.id = e2.managerId
group by e1.id
having count(*)>=58. 602. 好友申请 II ：谁有最多的好友
with用法
# Write your MySQL query statement below
with t1 as (
select requester_id as id1,accepter_id as id2
from RequestAccepted
union all
select accepter_id as id1,requester_id as id2
from RequestAccepted
)select id1 as id,count(distinct id2) as num
from t1
group by 1
order by 2 desc
limit 1
9. 608. 树节点
case
selectid,case when p_id is null then "Root"when id not in (select ifnull(p_id,'s') from Tree) then "Leaf"else "Inner"end as type
fromTree
)
