因是个人总结，只列出对自己有用的或较难的：
下面这道题，第一次拿到，我尝试用 开窗函数 ROW_NUMBER()OVER() 编号，但是发现不能够处理好连续的问题，
上网查找了别人的解法记录下来，其实原理 是 把 Logs 看成 三张表，每张表之间关联 -1 ，然后筛选出 Num 相等的
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 |
+-----------------+
select distinct log1.Num from Logs log1 join Logs log2 on log1.Id = log2.Id-1
join Logs log3 on log2.Id = log3.Id -1
where log1.Num = log2.Num and log2.Num = log3.Num
-- 同样的 连续问题
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 (date)、 人流量 (people)。
请编写一个查询语句，找出高峰期时段，要求连续三天及以上，并且每天人流量均不少于100。
例如，表 stadium：
+------+------------+-----------+
| id | 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 | date | people |
+------+------------+-----------+
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-08 | 188 |
+------+------------+-----------+
Note:
每天只有一行记录，日期随着 id 的增加而增加。
;WITH stadium(id,date,people) AS
(
SELECT 1,' 2017-01-01 ',10 union all
SELECT 2,' 2017-01-02 ',109 union all
SELECT 3,' 2017-01-03 ',150 union all
SELECT 4,' 2017-01-04 ',99 union all
SELECT 5,' 2017-01-05 ',145 union all
SELECT 6,' 2017-01-06 ',1455 union all
SELECT 7,' 2017-01-07 ',199 union all
SELECT 8,' 2017-01-08 ',188
)
, temp as(
select * from stadium where people > 100 --- 这里 可以避免后面写一大段 筛选条件
)
,tab as(
select id
,date
,people
,(id - (select max(id) from temp t where t.id < tmp.id)) as id_diff
from temp tmp
)
,final_tab as (
select t2.* from tab t1 join tab t2 on t1.id+1 = t2.id
join tab t3 on t2.id+1 = t3.id
where t1.id_diff = 1 and t2.id_diff = 1 and t3.id_diff = 1
)
select s.* from stadium s join final_tab f on s.id+1 = f.id
union ALL
select id
,date
,people from final_tab
union ALL
select s.* from stadium s join final_tab f on s.id-1 = f.id
leetcode 标记难度 困难
需求：
-- 部门工资前三高的员工
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 |
| 5 | Janet | 69000 | 1 |
| 6 | Randy | 85000 | 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，找出每个部门工资前三高的员工。例如，根据上述给定的表格，查询结果应返回：
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| IT | Randy | 85000 |
| IT | Joe | 70000 |
| Sales | Henry | 80000 |
| Sales | Sam | 60000 |
+------------+----------+--------+
第一种解法：
;with Employee(Id,Name,Salary,DepartmentId) AS
(
select 1,'Joe','70000',1 union all
select 2,'Henry','80000',2 union all
select 3,'Sam','60000',2 union all
select 4,'Max','90000',1 union all
select 5,'Janet','69000',1 union all
select 6,'Randy','85000',1
)
, Department(Id,Name) AS(
SELECT 1,'IT'
UNION ALL
SELECT 2,'Sales'
)
select d.Name as Department,e.Name as Employee,e.Salary from
(
SELECT *,ROW_NUMBER()OVER(partition by DepartmentId order by Salary desc) as Rank FROM Employee) e
join Department d on e.DepartmentId = d.Id
where Rank<=3
order by d.Id ASC
第二种解法：
为了避免 有相同排名出现，采用 DENSE_RANK 密级排名
select d.Name as Department,e.Name as Employee,e.Salary from
(
SELECT *,Dense_Rank()OVER(partition by DepartmentId order by Salary desc) as Rank FROM Employee) e
join Department d on e.DepartmentId = d.Id
where Rank<=3
order by d.Id ASC
leetcode 难度标记： 简单
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 |
+----+
select Id from Weather w where Temperature > (select Temperature from Weather l where dateadd(day,1,l.RecordDate) = w.RecordDate)
leetcode 难度标记： 简单
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 |
+---------------------+
解法：
SELECT MAX(SALARY) AS SecondHighestSalary FROM Employee WHERE Salary <(SELECT MAX(SALARY) FROM Employee)
leetcode 难度标记： 中等
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 T-SQL query statement below */
select d.Name as Department
,e.Name as Employee
,e.Salary as Salary
from Employee e
join Department d on e.DepartmentId = d.Id
join( SELECT DepartmentId
,max(Salary) as Salary
FROM Employee Group by DepartmentId
)
grp on e.Salary = grp.Salary
and e.DepartmentId = grp.DepartmentId
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 |
+----+------------------+
--- 这道题 leetcode 只支持 mysql 语法，很是恶心到我了。。。最后也不知道 运行成功了没有，但是 暂且记下 我的答案吧
# Write your MySQL query statement below
delete from Person where id not in(select id from ( select min(Id) as id
from Person
where Email in (select Email
from Person
group by Email
having count(Email) > 1
) ) as n )
and email in (select email from (select Email
from Person
group by Email
having count(Email)>1) as m)
leetcode 难度标记：中等
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。
;with seat(id,student) as(
SELECT 1 ,'Abbot' union all
SELECT 2 ,'Doris' union all
SELECT 3 ,'Emerson' union all
SELECT 4 ,'Green' union all
SELECT 5 ,'Jeames'
)
,odd_tab as(
select * from seat where id %2 = 1
)
,even_tab as(
select * from seat where id %2 = 0
)
select o.id,isnull(e.student,o.student) as student from odd_tab o left join even_tab e on o.id +1 = e.id
union ALL
select e.id,o.student from odd_tab o join even_tab e on o.id +1 = e.id
order by id asc
leetcode 难度标记 困难，个人觉得不该标记为困难
Trips 表中存所有出租车的行程信息。每段行程有唯一健 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。
+------------+-------------------+
| Day | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 | 0.33 |
| 2013-10-02 | 0.00 |
| 2013-10-03 | 0.50 |
+------------+-------------------+
Create table Trips (Id int,Client_Id int,
Driver_Id int, City_Id int,
Status varchar(50),
Request_at varchar(50));
Create table Users (Users_Id int,Banned varchar(50), Role varchar(50));
Truncate table Trips;
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('1', '1', '10', '1', 'completed','2013-10-01');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('2', '2', '11', '1','cancelled_by_driver', '2013-10-01');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('3', '3', '12', '6', 'completed','2013-10-01');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('4', '4', '13', '6','cancelled_by_client', '2013-10-01');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('5', '1', '10', '1', 'completed','2013-10-02');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('6', '2', '11', '6', 'completed','2013-10-02');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('7', '3', '12', '6', 'completed','2013-10-02');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('8', '2', '12', '12', 'completed','2013-10-03');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('9', '3', '10', '12', 'completed','2013-10-03');
insert into Trips (Id, Client_Id, Driver_Id,City_Id, Status, Request_at) values ('10', '4', '13', '12','cancelled_by_driver', '2013-10-03');
Truncate table Users;
insert into Users (Users_Id, Banned, Role)values ('1', 'No', 'client');
insert into Users (Users_Id, Banned, Role)values ('2', 'Yes', 'client');
insert into Users (Users_Id, Banned, Role)values ('3', 'No', 'client');
insert into Users (Users_Id, Banned, Role)values ('4', 'No', 'client');
insert into Users (Users_Id, Banned, Role)values ('10', 'No', 'driver');
insert into Users (Users_Id, Banned, Role)values ('11', 'No', 'driver');
insert into Users (Users_Id, Banned, Role)values ('12', 'No', 'driver');
insert into Users (Users_Id, Banned, Role)values ('13', 'No', 'driver');
select t.Request_at as [Day],Round((sum(case when Status like '%cancelled%' then 1 else 0 end) * 1.0 / count(status) ),2) as [Cancellation Rate]
from Trips t join USERs u
on t.client_id = u.users_id
where Request_at between '2013-10-01' and '2013-10-03'
and u.Role = 'client' and Banned = 'No'
group by Request_at
小结：
目前，leetcode 上的 sql 题均已刷完，感觉难度一般，个人卡壳的地方 是 连续问题的 求解， 里边有两道 连续问题，本文第一道和第二道
下次见到 连续问题应该首先想 多表 id+1 ， id+2 ,...,id+n 这种方式求解，而不应想编号，编号不足以解决连续问题