Task06：综合练习

练习一: 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。

创建Department 表，包含公司所有部门的信息。

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。

create table Employee(
	Id int not null auto_increment,
    Name varchar(8) not null,
    Salary int not null,
    Departmentid int not null,
    primary key (Id));

insert into Employee (Name, Salary, Department) values ('Joe', 70000, 1);
insert into Employee (Name, Salary, Department) values ('Henry', 80000, 2);
insert into Employee (Name, Salary, Department) values ('Sam', 60000, 2);
insert into Employee (Name, Salary, Department) values ('Max', 90000, 1);

create table Department(
	Id int not null auto_increment,
    Name varchar(8) not null,
    primary key (Id));
    
insert into Department (Name) values ('IT');
insert into Department (Name) values ('Sales');


select d.Name as Department, e.Name as Employee, e.Salary as Salary
from Department as d
join (select Name, Salary, DepartmentId from Employee as e1
	  where Salary = (select max(Salary) from Employee as e2
					  where e1.DepartmentId = e2.DepartmentId
                      group by e2.DepartmentId)) as e
on d.Id = e.DepartmentId
order by e.Salary desc;

练习二: 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：

假如数据输入的是上表，则输出结果如下：

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。

create table seat(
	id int not null auto_increment,
    student varchar(8) not null,
    primary key (id));
    
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');

select (case when s.id % 2 = 0 then s.id - 1
			 when s.id % 2 != 0 and s.id != c.counts then s.id + 1
             else s.id end) as id,
		student
from seat as s, (select count(*) as counts from seat) as c
order by id;

练习三: 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：

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）：

create table scores(
	Id int not null auto_increment,
    Score float not null,
    primary key(Id));

insert into scores (Score) values (3.50);
insert into scores (Score) values (3.65);
insert into scores (Score) values (4.00);
insert into scores (Score) values (3.85);
insert into scores (Score) values (4.00);
insert into scores (Score) values (3.65);

-- 使用开窗函数
select Score, dense_rank() over(order by Score desc) as 'rank' 
from scores;

-- 使用关联子查询
select s1.Score, 
	   (select count(distinct s2.Score) from scores as s2
       where s2.Score >= s1.Score) as 'rank'
from scores as s1
order by s1.Score desc;

练习四：连续出现的数字（难度：中等）

编写一个 SQL 查询，查找所有至少连续出现三次的数字。

例如，给定上面的 Logs 表， 1 是唯一连续出现至少三次的数字。

create table Logs(
	Id int not null auto_increment,
    Num int not null,
    primary key(Id));

insert into Logs (Num) values (1);
insert into Logs (Num) values (1);
insert into Logs (Num) values (1);
insert into Logs (Num) values (2);
insert into Logs (Num) values (1);
insert into Logs (Num) values (2);
insert into Logs (Num) values (2);


select distinct s1.num as ConsecutiveNums from 
	(select *, row_number() over(order by id) - row_number() over(partition by num order by id) as uni
	from logs) as s1
group by s1.num, s1.uni
having count(*) >= 3;

练习五：树节点 （难度：中等）

对于tree表，id是树节点的标识，p_id是其父节点的id。

每个节点都是以下三种类型中的一种：

• Root: 如果节点是根节点。

• Leaf: 如果节点是叶子节点。

• Inner: 如果节点既不是根节点也不是叶子节点。
  写一条查询语句打印节点id及对应的节点类型。按照节点id排序。上面例子的对应结果为：
  
  说明

• 节点’1’是根节点，因为它的父节点为NULL，有’2’和’3’两个子节点。

• 节点’2’是内部节点，因为它的父节点是’1’，有子节点’4’和’5’。

• 节点’3’，‘4’，'5’是叶子节点，因为它们有父节点但没有子节点。
  下面是树的图形：
  
  注意

如果一个树只有一个节点，只需要输出根节点属性。

create table tree(
	id int not null auto_increment,
    p_id int,
    primary key (id));
    
insert into tree (p_id) values (null);
insert into tree (p_id) values (1);
insert into tree (p_id) values (1);
insert into tree (p_id) values (2);
insert into tree (p_id) values (2);

-- 使用case when
select id, (case when isnull(p_id) then 'Root'
				 when id in (select p_id from tree) then 'Inner'
                 else 'Leaf' end) as Type
from tree;

-- 使用if
select id, if(isnull(p_id), 'Root', 
			if(id in (select p_id from tree), 'Inner', 'Leaf')) as Type
from tree;

练习六：至少有五名直接下属的经理 （难度：中等）

Employee表包含所有员工及其上级的信息。每位员工都有一个Id，并且还有一个对应主管的Id（ManagerId）。

针对Employee表，写一条SQL语句找出有5个下属的主管。对于上面的表，结果应输出：

注意:

没有人向自己汇报。

create table Employee1(
	Id int not null,
    Name varchar(8) not null,
    Department char(1) not null,
    ManagerId int,
    primary key (Id));
    
insert into Employee1 values (101, 'John', 'A', null);
insert into Employee1 values (102, 'Dan', 'A', 101);
insert into Employee1 values (103, 'James', 'A', 101);
insert into Employee1 values (104, 'Amy', 'A', 101);
insert into Employee1 values (105, 'Anne', 'A', 101);
insert into Employee1 values (106, 'Ron', 'B', 101);

select Name from Employee1
where Id in (select ManagerId from Employee1
			 group by ManagerId
             having count(*) >= 5);

练习七: 分数排名 （难度：中等）

练习三的分数表，实现排名功能，但是排名需要是非连续的，如下：

select Score, rank() over(order by Score desc) as 'rank' from scores;

练习八：查询回答率最高的问题 （难度：中等）

求出survey_log表中回答率最高的问题，表格的字段有：uid, action, question_id, answer_id, q_num, timestamp。

uid是用户id；action的值为：“show”， “answer”， “skip”；当action是"answer"时，answer_id不为空，相反，当action是"show"和"skip"时为空（null）；q_num是问题的数字序号。

写一条sql语句找出回答率最高的问题。
举例：

输入

输出

问题285的回答率为1/1，然而问题369的回答率是0/1，所以输出是285。

注意：最高回答率的意思是：同一个问题出现的次数中回答的比例。

create table survey_log(
	uid int not null,
    action varchar(8) not null,
    question_id int not null,
    answer_id int,
    q_num int not null,
    timestamp int not null,
    primary key(timestamp));

insert into survey_log values (5, 'show', 285, null, 1, 123);
insert into survey_log values (5, 'answer', 285, 124124, 1, 124);
insert into survey_log values (5, 'show', 369, null, 2, 125);
insert into survey_log values (5, 'skip', 369, null, 2, 126);

select question_id as survey_log
from (select question_id, 
			sum(case when action = 'answer' then 1 else 0 end) / sum(case when action = 'show' then 1 else 0 end) as ratio
	from survey_log
	group by question_id) as s
order by ratio desc
limit 1;

练习九：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）：

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：

此外，请考虑实现各部门前N高工资的员工功能。

select d.Name as Department, e.Name as Employee, e.Salary
from (select e1.DepartmentId, e1.Name, e1.Salary
		from (select *, rank() over (partition by DepartmentId order by Salary desc) as 'rank'
			from employee) as e1
		where e1.rank <= 3) as e
join Department as d
on e.DepartmentId = d.Id;

练习十：平面上最近距离 (难度: 困难）

point_2d表包含一个平面内一些点（超过两个）的坐标值（x，y）。

写一条查询语句求出这些点中的最短距离并保留2位小数。

最短距离是1，从点（-1，-1）到点（-1，2）。所以输出结果为：

| shortest |

1.00

注意：所有点的最大距离小于10000。

create table point_2d (
	x int not null,
    y int not null,
    primary key (x, y));
    
insert into point_2d values (-1, -1);
insert into point_2d values (0, 0);
insert into point_2d values (-1, -2);

select round(min(sqrt(power(p1.x - p2.x, 2) + power(p1.y - p2.y, 2))), 2) as shortest
from point_2d as p1, point_2d as p2
where p1.x != p2.x or p1.y != p2.y;

-- 效率高
select round(min(sqrt(power(p1.x - p2.x, 2) + power(p1.y - p2.y, 2))), 2) as shortest
from point_2d as p1, point_2d as p2
where p1.x < p2.x or (p1.x = p2.x and p1.y < p2.y);

练习十一：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’)。

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。

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。

create table Trips(
	Id int not null auto_increment,
    Client_Id int not null,
    Driver_Id int not null,
    City_Id int not null,
    Status enum('completed','cancelled_by_driver','cancelled_by_client') not null,
    Request_at date not null,
    primary key(Id)
    );
    
insert into Trips values (1, 1, 10, 1, 'completed', '2013-10-1');
insert into Trips values (2, 2, 11, 1, 'cancelled_by_driver', '2013-10-1');
insert into Trips values (3, 3, 12, 6, 'completed', '2013-10-1');
insert into Trips values (4, 4, 13, 6, 'cancelled_by_client', '2013-10-1');
insert into Trips values (5, 1, 10, 1, 'completed', '2013-10-2');
insert into Trips values (6, 2, 11, 6, 'completed', '2013-10-2');
insert into Trips values (7, 3, 12, 6, 'completed', '2013-10-2');
insert into Trips values (8, 2, 12, 12, 'completed', '2013-10-3');
insert into Trips values (9, 3, 10, 12, 'completed', '2013-10-3');
insert into Trips values (10, 4, 13, 12, 'cancelled_by_driver', '2013-10-3');

create table Users(
	Users_Id int not null,
    Banned enum('Yes', 'No') not null,
    Role enum('client', 'driver') not null,
    primary key(Users_Id)
    );
    
insert into Users values (1, 'No', 'client');
insert into Users values (2, 'Yes', 'client');
insert into Users values (3, 'No', 'client');
insert into Users values (4, 'No', 'client');
insert into Users values (10, 'No', 'driver');
insert into Users values (11, 'No', 'driver');
insert into Users values (12, 'No', 'driver');
insert into Users values (13, 'No', 'driver');

select Request_at as Day,
round(sum(case when Status not like 'completed' then 1 else 0 end) / count(*), 2) as 'Cancellation Rate'
from (select * from Trips
	  where Client_Id in (select Users_Id 
						  from Users
                          where Banned like 'No')
	  and Driver_Id in (select Users_Id 
						  from Users
                          where Banned like 'No')) as s
where Request_at between '2013-10-01' and '2013-10-03'
group by Request_at;