1.
项目七: 各部门工资最高的员工（难度：中等）
创建 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 |
+----+-------+--------+--------------+
创建 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 |
+------------+----------+--------+
1
create table Employee(
ID INT auto_increment primary key,
NAME VARCHAR(50) NOT NULL,
SALARY INT NOT NULL,
DEPARTMENT INT NOT NULL
);
insert into Emploee (NAME,SALARY,Departmen) values(Joe,70000,1);
insert into Emploee(Name,Salary,Department) values(Henry,80000,2);
insert into Emploee(Name,Salary,Department) values(Sam,60000,2);
insert into Emploee(Name,Salary,Department) values(Max,90000,1);
2.
create table Department(
Id int primary key auto_increment,
Name varchar not null
);
insert into Department(Name) values (IT);
insert into Department(Name) values (Sales);
3
select Department.Name as Department,
Emploe.Name as Emploee,
max(Emploee.Salary) as Salary from Emploee,
Department where Employee.Id=Department.Id group by Emploee.Department;
2.项目八: 换座位（难度：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？
请创建如下所示 seat 表：
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 |
+---------+---------+
1.
create table seat(
id int primary key auto_increment,
student varchar not null
);
insert into seat(student) values(Abbot);
insert into seat(student) values(Doris);
insert into seat(student) values(Emerson);
insert into seat(student) values(Green);
insert into seat(student) values(Jeames);
2.
UPDATE seat s1
JOIN seat s2
ON (s1.id % 2 = 1 AND s2.id = s1.id+1)
SET s1.student=s2.student,s2.student=s1.studentWHERE s1.id+1 <> null;
3.
项目九: 分数排名（难度：中等）
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”。
创建以下 score 表：
+----+-------+
| 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 |
+-------+------+
1.
create table score(
Id int auto_increment primary key,
Score float not null
);
2.
SELECT Score,
CASE
WHEN @prevRank = Score THEN @curRank
WHEN @prevRank := Score THEN @curRank := @curRank + 1
END AS Rank
FROM scores,
(SELECT @curRank :=0, @prevRank := NULL)
ORDER BY Score desc;
4.
项目十：行程和用户（难度：困难）
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。
+------------+-------------------+
| Day | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 | 0.33 |
| 2013-10-02 | 0.00 |
| 2013-10-03 | 0.50 |
+------------+-------------------+
select t1.Request_at,t1.c1/t2.c2 as Cancellation Rate from
(select count(Status) as c1 Request_at from Trips where Client_Id in (select Users_id from Users where Role=client && Banned=No;) and Status <> completed group by Request_at;) t1,
(select count(Status) as c2 Request_at from Trips where Client_Id in (select Users_id from Users where Role=client && Banned=No;) group by Request_at;) t2
where
t1.Request_at=t2.Request_at;
5.
项目十一：各部门前3高工资的员工（难度：中等）
将项目7中的 employee 表清空，重新插入以下数据（其实是多插入5,6两行）：
+----+-------+--------+--------------+
| 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 |
+----+-------+--------+--------------+
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 |
+------------+----------+--------+
此外，请考虑实现各部门前N高工资的员工功能。
select Department.Name as Department,EmployeeName as Employee,Employee.Salary from
Department join Employee on
order by Employee.salary
desc
limit 3
group by
Department.name
where
Employee.Department_Id=Department.Id;
6.
项目十二 分数排名 - （难度：中等）
依然是昨天的分数表，实现排名功能，但是排名是非连续的，如下：
+-------+------+
| Score | Rank |
+-------+------+
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 3 |
| 3.65 | 4 |
| 3.65 | 4 |
| 3.50 | 6 |
+-------+------
SELECT Score Rank FROM
(SELECT Score,
@curRank := IF(@prevRank = Score, @curRank, @incRank) AS Rank,
@incRank := @incRank + 1,
@prevRank := Score
FROM scores,(
SELECT @curRank :=0, @prevRank := NULL, @incRank := 1
)
ORDER BY Score desc;