【2020-02-25】leetcode sql

177、第二高薪水
编写一个 SQL 查询，获取 Employee 表中第二高的薪水（Salary） 。

+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
例如上述 Employee 表，SQL查询应该返回 200 作为第二高的薪水。如果不存在第二高的薪水，那么查询应返回 null。

+---------------------+
| SecondHighestSalary |
+---------------------+
| 200 |
+---------------------+

# Write your MySQL query statement below
SELECT
    (SELECT DISTINCT
            Salary
        FROM
            Employee
        ORDER BY Salary DESC
        LIMIT 1 OFFSET 1) AS SecondHighestSalary

177、第N高薪水
编写一个 SQL 查询，获取 Employee 表中第 n 高的薪水（Salary）。

+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
例如上述 Employee 表，n = 2 时，应返回第二高的薪水 200。如果不存在第 n 高的薪水，那么查询应返回 null。

+------------------------+
| getNthHighestSalary(2) |
+------------------------+
| 200 |
+------------------------+

CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
set n=N-1;
  RETURN (
      # Write your MySQL query statement below.
      select distinct salary
      from Employee
      order by salary desc limit n,1

  );
END

178、分数排名
编写一个 SQL 查询来实现分数排名。如果两个分数相同，则两个分数排名（Rank）相同。请注意，平分后的下一个名次应该是下一个连续的整数值。换句话说，名次之间不应该有“间隔”。

+----+-------+
| Id | Score |
+----+-------+
| 1 | 3.50 |
| 2 | 3.65 |
| 3 | 4.00 |
| 4 | 3.85 |
| 5 | 4.00 |
| 6 | 3.65 |
+----+-------+
例如，根据上述给定的 Scores 表，你的查询应该返回（按分数从高到低排列）：

+-------+------+
| Score | Rank |
+-------+------+
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 2 |
| 3.65 | 3 |
| 3.65 | 3 |
| 3.50 | 4 |
+-------+------+

在mysql 实现Rank高级排名函数https://www.cnblogs.com/caicaizi/p/9803013.html
要在mysql中声明一个变量，你必须在变量名之前使用@符号。FROM子句中的(@curRank := 0)部分允许我们进行变量初始化，而不需要单独的SET命令。也可以使用SET。

#在Score不为0的情况下，@pre := Score为真，
# Write your MySQL query statement below
SELECT Score, CASE
WHEN @pre = Score THEN @rk + 0
WHEN @pre := Score THEN @rk := @rk + 1
ELSE @rk := @rk + 1
END AS Rank
FROM Scores, (SELECT @pre := NULL, @rk := 0) AS init
ORDER BY Score DESC;

180、连续出现的数字
编写一个 SQL 查询，查找所有至少连续出现三次的数字。

+----+-----+
| Id | Num |
+----+-----+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
| 5 | 1 |
| 6 | 2 |
| 7 | 2 |
+----+-----+
例如，给定上面的 Logs 表， 1 是唯一连续出现至少三次的数字。

+-----------------+
| ConsecutiveNums |
+-----------------+
| 1 |
+-----------------+

# Write your MySQL query statement below
select distinct  dd.Num ConsecutiveNums
from  (
select d.Num,
       @n :=if(@pre=Num,@n+1,@n:=1) count,
       @pre:=Num
from Logs d,
     (select @pre:=null, @n :=1)r)dd
where dd.count>=3;

184、部门工资最高的员工
Employee 表包含所有员工信息，每个员工有其对应的 Id, salary 和 department Id。

+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
+----+-------+--------+--------------+
Department 表包含公司所有部门的信息。

+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
编写一个 SQL 查询，找出每个部门工资最高的员工。例如，根据上述给定的表格，Max 在 IT 部门有最高工资，Henry 在 Sales 部门有最高工资。

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+

# Write your MySQL query statement below
SELECT
    Department.name AS 'Department',
    Employee.name AS 'Employee',
    Salary
FROM
    Employee
        JOIN
    Department ON Employee.DepartmentId = Department.Id
WHERE
    (Employee.DepartmentId , Salary) IN
    (   SELECT
            DepartmentId, MAX(Salary)
        FROM
            Employee
        GROUP BY DepartmentId
    )
;

185、部门工资前三高的所有员工
Employee 表包含所有员工信息，每个员工有其对应的工号 Id，姓名 Name，工资 Salary 和部门编号 DepartmentId 。

+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 85000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
| 5 | Janet | 69000 | 1 |
| 6 | Randy | 85000 | 1 |
| 7 | Will | 70000 | 1 |
+----+-------+--------+--------------+
Department 表包含公司所有部门的信息。

+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
编写一个 SQL 查询，找出每个部门获得前三高工资的所有员工。例如，根据上述给定的表，查询结果应返回：

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Randy | 85000 |
| IT | Joe | 85000 |
| IT | Will | 70000 |
| Sales | Henry | 80000 |
| Sales | Sam | 60000 |
+------------+----------+--------+
解释：

IT 部门中，Max 获得了最高的工资，Randy 和 Joe 都拿到了第二高的工资，Will 的工资排第三。销售部门（Sales）只有两名员工，Henry 的工资最高，Sam 的工资排第二。

# Write your MySQL query statement below

SELECT
    Department.NAME AS Department,
    e1.NAME AS Employee,
    e1.Salary AS Salary 
FROM
    Employee AS e1,Department 
WHERE
    e1.DepartmentId = Department.Id 
    AND 3 > (SELECT  count( DISTINCT e2.Salary ) 
             FROM   Employee AS e2 
             WHERE  e1.Salary < e2.Salary   AND e1.DepartmentId = e2.DepartmentId   ) 
ORDER BY Department.NAME,Salary DESC;

196、删除重复的电子邮箱
编写一个 SQL 查询，来删除 Person 表中所有重复的电子邮箱，重复的邮箱里只保留 Id 最小 的那个。

+----+------------------+
| Id | Email |
+----+------------------+
| 1 | john@example.com |
| 2 | bob@example.com |
| 3 | john@example.com |
+----+------------------+
Id 是这个表的主键。
例如，在运行你的查询语句之后，上面的 Person 表应返回以下几行:

+----+------------------+
| Id | Email |
+----+------------------+
| 1 | john@example.com |
| 2 | bob@example.com |
+----+------------------+

提示：

执行 SQL 之后，输出是整个 Person 表。
使用 delete 语句。

# Write your MySQL query statement below
DELETE FROM Person
WHERE Id NOT IN (   -- 删除不在查询结果中的值
    SELECT id FROM
   (
       SELECT MIN(Id) AS Id -- 排除Email相同时中Id较大的行
       FROM Person
       GROUP BY Email
   ) AS temp    -- 此处需使用临时表，否则会发生报错
)

197、上升的温度
给定一个 Weather 表，编写一个 SQL 查询，来查找与之前（昨天的）日期相比温度更高的所有日期的 Id。

+---------+------------------+------------------+
| Id(INT) | RecordDate(DATE) | Temperature(INT) |
+---------+------------------+------------------+
| 1 | 2015-01-01 | 10 |
| 2 | 2015-01-02 | 25 |
| 3 | 2015-01-03 | 20 |
| 4 | 2015-01-04 | 30 |
+---------+------------------+------------------+
例如，根据上述给定的 Weather 表格，返回如下 Id:

+----+
| Id |
+----+
| 2 |
| 4 |
+----+

# 如果反过来【用b的日期-a的日期，a的温度>b.温度，输出a的ID】，结果就会变少？为什么
# Write your MySQL query statement below
select b.Id
from Weather a join Weather b
where DATEDIFF(a.RecordDate,b.RecordDate)=-1
and a.Temperature<b.Temperature

262、行程和用户
Trips 表中存所有出租车的行程信息。每段行程有唯一键 Id，Client_Id 和 Driver_Id 是 Users 表中 Users_Id 的外键。Status 是枚举类型，枚举成员为 (‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’)。

+----+-----------+-----------+---------+--------------------+----------+
| Id | Client_Id | Driver_Id | City_Id | Status |Request_at|
+----+-----------+-----------+---------+--------------------+----------+
| 1 | 1 | 10 | 1 | completed |2013-10-01|
| 2 | 2 | 11 | 1 | cancelled_by_driver|2013-10-01|
| 3 | 3 | 12 | 6 | completed |2013-10-01|
| 4 | 4 | 13 | 6 | cancelled_by_client|2013-10-01|
| 5 | 1 | 10 | 1 | completed |2013-10-02|
| 6 | 2 | 11 | 6 | completed |2013-10-02|
| 7 | 3 | 12 | 6 | completed |2013-10-02|
| 8 | 2 | 12 | 12 | completed |2013-10-03|
| 9 | 3 | 10 | 12 | completed |2013-10-03|
| 10 | 4 | 13 | 12 | cancelled_by_driver|2013-10-03|
+----+-----------+-----------+---------+--------------------+----------+
Users 表存所有用户。每个用户有唯一键 Users_Id。Banned 表示这个用户是否被禁止，Role 则是一个表示（‘client’, ‘driver’, ‘partner’）的枚举类型。

+----------+--------+--------+
| Users_Id | Banned | Role |
+----------+--------+--------+
| 1 | No | client |
| 2 | Yes | client |
| 3 | No | client |
| 4 | No | client |
| 10 | No | driver |
| 11 | No | driver |
| 12 | No | driver |
| 13 | No | driver |
+----------+--------+--------+
写一段 SQL 语句查出 2013年10月1日 至 2013年10月3日 期间非禁止用户的取消率。基于上表，你的 SQL 语句应返回如下结果，取消率（Cancellation Rate）保留两位小数。

取消率的计算方式如下：(被司机或乘客取消的非禁止用户生成的订单数量) / (非禁止用户生成的订单总数)

+------------+-------------------+
| Day | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 | 0.33 |
| 2013-10-02 | 0.00 |
| 2013-10-03 | 0.50 |
+------------+-------------------+

# Write your MySQL query statement below
select temp.request_at Day,
round(sum(case temp.status when 'completed' then 0 else 1 end)/count(temp.status),2) 'Cancellation Rate' 
from (select status,request_at from trips t left join users u on t.client_id = u.users_id where u.banned ='no') temp 
where request_at between '2013-10-01' and '2013-10-03' 
group by temp.request_at

601、体育馆人流量
X 市建了一个新的体育馆，每日人流量信息被记录在这三列信息中：序号 (id)、日期 (visit_date)、 人流量 (people)。

请编写一个查询语句，找出人流量的高峰期。高峰期时，至少连续三行记录中的人流量不少于100。

例如，表 stadium：

+------+------------+-----------+
| id | visit_date | people |
+------+------------+-----------+
| 1 | 2017-01-01 | 10 |
| 2 | 2017-01-02 | 109 |
| 3 | 2017-01-03 | 150 |
| 4 | 2017-01-04 | 99 |
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-08 | 188 |
+------+------------+-----------+
对于上面的示例数据，输出为：

+------+------------+-----------+
| id | visit_date | people |
+------+------------+-----------+
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-08 | 188 |
+------+------------+-----------+

# Write your MySQL query statement below
SELECT id, visit_date, people
FROM (
    SELECT r1.*, @flag := if((r1.countt >= 3 OR @flag = 1) AND r1.countt != 0, 1, 0) AS flag
    FROM (
        SELECT s.*, @count := if(s.people >= 100, @count + 1, 0) AS `countt`
        FROM stadium s, (SELECT @count := 0) b
    ) r1, (SELECT @flag := 0) c
    ORDER BY id DESC
) result
WHERE flag = 1 ORDER BY id;

626、换座位
小美是一所中学的信息科技老师，她有一张 seat 座位表，平时用来储存学生名字和与他们相对应的座位 id。

其中纵列的 id 是连续递增的

小美想改变相邻俩学生的座位。

你能不能帮她写一个 SQL query 来输出小美想要的结果呢？

示例：

+---------+---------+
| id | student |
+---------+---------+
| 1 | Abbot |
| 2 | Doris |
| 3 | Emerson |
| 4 | Green |
| 5 | Jeames |
+---------+---------+
假如数据输入的是上表，则输出结果如下：

+---------+---------+
| id | student |
+---------+---------+
| 1 | Doris |
| 2 | Abbot |
| 3 | Green |
| 4 | Emerson |
| 5 | Jeames |
+---------+---------+
注意：

如果学生人数是奇数，则不需要改变最后一个同学的座位。

# Write your MySQL query statement below
SELECT
    s1.id, COALESCE(s2.student, s1.student) AS student
FROM
    seat s1
        LEFT JOIN
    seat s2 ON ((s1.id + 1) ^ 1) - 1 = s2.id
ORDER BY s1.id;