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。
+----+-------+--------+--------------+
| 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 |
+------------+----------+--------+
DROP TABLE if EXISTS Employee;
CREATE TABLE Employee
(id INT,
name VARCHAR(20),
salary INT,
departmentid INT,
PRIMARY KEY (id));
INSERT INTO Employee VALUES(1,'Joe',70000,1);
INSERT INTO Employee VALUES(2,'Henry',80000,2);
INSERT INTO Employee VALUES(3,'Sam',60000,2);
INSERT INTO Employee VALUES(4,'Max',90000,1);
SELECT * FROM Employee;
DROP TABLE if EXISTS Department;
CREATE TABLE Department
(id INT,
name VARCHAR(20),
PRIMARY KEY (id));
INSERT INTO Department VALUES(1,'IT');
INSERT INTO Department VALUES(2,'Sales');
SELECT * FROM Department;
SELECT
t.Department,
ee. NAME AS Employee,
t.Salary
FROM
(
SELECT
d. NAME AS Department,
e.departmentid,
MAX(e.salary) AS Salary
FROM
Employee e
LEFT JOIN Department d ON e.departmentid = d.id
GROUP BY
d. NAME
) t
LEFT JOIN Employee ee ON t.departmentid = ee.departmentid
AND t.Salary = ee.salary
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：
+---------+---------+
| 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。
DROP TABLE if exists seat;
CREATE TABLE seat
(id INT,
student VARCHAR(20),
PRIMARY KEY (id));
INSERT INTO seat VALUES(1,'Abbot');
INSERT INTO seat VALUES(2,'Doris');
INSERT INTO seat VALUES(3,'Emerson');
INSERT INTO seat VALUES(4,'Green');
INSERT INTO seat VALUES(5,'Jeames');
SELECT * FROM seat;
SELECT
CASE WHEN id = (SELECT MAX(id) FROM seat) THEN id
WHEN MOD(id,2)=1 THEN id+1
WHEN MOD(id,2)=0 THEN id-1
ELSE NULL END AS id,
student
FROM seat
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：
+----+-------+
| 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 |
+-------+------+
DROP TABLE if exists score;
CREATE TABLE score
(id INT,
score FLOAT(3,2),
PRIMARY KEY (id));
INSERT INTO score VALUES(1,3.50);
INSERT INTO score VALUES(2,3.65);
INSERT INTO score VALUES(3,4.00);
INSERT INTO score VALUES(4,3.85);
INSERT INTO score VALUES(5,4.00);
INSERT INTO score VALUES(6,3.65);
SELECT * FROM score;
-- 考察窗口函数
-- PARTITION BY 用于分组，不分组时，省略即可
-- ORDER BY 用于排序，默认升级，降序使用 DESC关键字
-- RANK() 美式排名，跳过式排序，得分相同时排序相同
-- DENSE_RANK() 中式排名， 递增式排序，得分相同时排序相同
-- ROW_NUMBER() 赋予唯一连续的名次
SELECT
score,
DENSE_RANK() OVER (ORDER BY score DESC) AS rank1
FROM score;
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。
+----+-----+
| 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 |
+-----------------+
DROP TABLE if exists logs;
CREATE TABLE logs
(id INT,
num INT,
PRIMARY KEY (id));
INSERT INTO logs VALUES (1, 1);
INSERT INTO logs VALUES (2, 1);
INSERT INTO logs VALUES (3, 1);
INSERT INTO logs VALUES (4, 2);
INSERT INTO logs VALUES (5, 1);
INSERT INTO logs VALUES (6, 2);
INSERT INTO logs VALUES (7, 2);
SELECT * FROM logs;
-- 连续出现的意味着相同数字的 Id 是连着的，由于这题问的是至少连续出现 3 次，我们使用 Logs 并检查是否有 3 个连续的相同数字。
-- 然后我们从上表中选择任意的 Num 获得想要的答案。同时我们需要添加关键字 DISTINCT ，因为如果一个数字连续出现超过 3 次，会返回重复元素。
SELECT DISTINCT
a.num AS ConsecutiveNums
FROM
LOGS a,
LOGS b,
LOGS c
WHERE
a.id = b.id - 1
AND b.id = c.id - 1
AND a.num = b.num
AND b.num = c.num;
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。
+----+------+
| id | p_id |
+----+------+
| 1 | null |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
| 5 | 2 |
+----+------+
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：
+----+------+
| id | Type |
+----+------+
| 1 | Root |
| 2 | Inner|
| 3 | Leaf |
| 4 | Leaf |
| 5 | Leaf |
+----+------+
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：
1
/ \
2 3
/ \
4 5
注意
如果一个树只有一个节点，只需要输出根节点属性。
DROP TABLE if exists tree;
CREATE TABLE tree
(id INT,
p_id INT,
PRIMARY KEY (id));
INSERT INTO tree VALUES (1, null);
INSERT INTO tree VALUES (2, 1);
INSERT INTO tree VALUES (3, 1);
INSERT INTO tree VALUES (4, 2);
INSERT INTO tree VALUES (5, 2);
SELECT * FROM tree;
SELECT id,
CASE WHEN p_id IS NULL THEN 'Root'
WHEN id IN (SELECT p_id FROM tree) THEN 'Inner'
ELSE 'Leaf' END 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）。
+------+----------+-----------+----------+
|Id |Name |Department |ManagerId |
+------+----------+-----------+----------+
|101 |John |A |null |
|102 |Dan |A |101 |
|103 |James |A |101 |
|104 |Amy |A |101 |
|105 |Anne |A |101 |
|106 |Ron |B |101 |
+------+----------+-----------+----------+
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：
+-------+
| Name |
+-------+
| John |
+-------+
注意:
没有人向自己汇报。
DROP TABLE if exists Employee2;
CREATE TABLE Employee2
(id INT,
name varchar(20),
department varchar(20),
managerid INT,
PRIMARY KEY (id));
INSERT INTO Employee2 VALUES (101, 'John', 'A', null);
INSERT INTO Employee2 VALUES (102, 'Dan', 'A', 101);
INSERT INTO Employee2 VALUES (103, 'James', 'A', 101);
INSERT INTO Employee2 VALUES (104, 'Amy', 'A', 101);
INSERT INTO Employee2 VALUES (105, 'Anne', 'A', 101);
INSERT INTO Employee2 VALUES (106, 'Ron', 'B', 101);
SELECT * FROM Employee2;
SELECT
e1. NAME
FROM Employee2 e1
JOIN (
SELECT managerid
FROM Employee2
GROUP BY managerid
HAVING count(id) = 5
) t1 ON e1.id = t1.managerid;
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：
+-------+------+
| Score | Rank |
+-------+------+
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 3 |
| 3.65 | 4 |
| 3.65 | 4 |
| 3.50 | 6 |
+-------+------
SELECT score,
RANK() OVER (ORDER BY score DESC) AS rank1
FROM score;
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
| uid | action | question_id | answer_id | q_num | timestamp |
|---|---|---|---|---|---|
| 5 | show | 285 | null | 1 | 123 |
| 5 | answer | 285 | 124124 | 1 | 124 |
| 5 | show | 369 | null | 2 | 125 |
| 5 | skip | 369 | null | 2 | 126 |
输出
| survey_log |
|---|
| 285 |
说明
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。
DROP TABLE IF EXISTS survey_log;
CREATE TABLE survey_log (
uid INT,
action VARCHAR (20),
question_id INT,
answer_id INT,
q_num INT,
TIMESTAMP INT
);
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 * FROM survey_log;
SELECT a.question_id
FROM (
SELECT question_id,
sum(CASE WHEN answer_id IS NOT NULL THEN 1 ELSE 0 END) / sum(CASE WHEN action = 'show' THEN 1 ELSE 0 END) AS ratio
FROM survey_log
GROUP BY question_id
ORDER BY ratio DESC
LIMIT 1) a;
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）：
+----+-------+--------+--------------+
| 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。
-- 建表语句
DROP TABLE IF EXISTS employee9;
CREATE TABLE employee9 SELECT * FROM employee;
SELECT * FROM employee9;
INSERT INTO employee9 VALUES(5, 'Janet', 69000, 1);
INSERT INTO employee9 VALUES(6, 'Randy', 85000, 1);
SELECT * FROM department;
-- 窗口函数解法
SELECT t.department, t.employee, t.salary
FROM(
SELECT d.name AS department, e.name AS employee, e.salary,
DENSE_RANK() OVER ( PARTITION BY d.name ORDER BY e.salary) AS rankn
FROM employee9 e
JOIN department d
ON e.departmentid = d.id) t
WHERE t.rankn <= 3
ORDER BY department, salary DESC;
-- 引入变量，实现各部门前N高工资的员工
SET @NN=3;
SELECT t.department, t.employee, t.salary
FROM(
SELECT d.name AS department, e.name AS employee, e.salary,
DENSE_RANK() OVER ( PARTITION BY d.name ORDER BY e.salary) AS rankn
FROM employee9 e
JOIN department d
ON e.departmentid = d.id) t
WHERE t.rankn <= @NN
ORDER BY department, salary DESC;
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。
|x | y |
|----|----|
| -1 | -1 |
| 0 | 0 |
| -1 | -2 |
最短距离是1，从点（-1，-1）到点（-1，-2）。所以输出结果为：
| shortest |
1.00
+--------+
|shortest|
+--------+
|1.00 |
+--------+
注意：所有点的最大距离小于10000。
-- 建表语句
DROP TABLE IF EXISTS point_2d;
CREATE TABLE point_2d (
x INT,
y INT
);
INSERT INTO point_2d VALUES(-1, -1);
INSERT INTO point_2d VALUES( 0, 0);
INSERT INTO point_2d VALUES(-1, -2);
SELECT * FROM point_2d;
SELECT
MIN(ROUND(POW(POW(ABS(p1.x-p2.x),2)+POW(ABS(p1.y-p2.y),2),0.5),2)) AS shortest
FROM point_2d p1
JOIN point_2d p2
ON p1.x!=p2.x
OR 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’)。
| Id | Client_Id | Driver_Id | City_Id | Status | Request_at |
|---|---|---|---|---|---|
| 1 | 1 | 10 | 1 | completed | 2013-10-1 |
| 2 | 2 | 11 | 1 | cancelled_by_driver | 2013-10-1 |
| 3 | 3 | 12 | 6 | completed | 2013-10-1 |
| 4 | 4 | 13 | 6 | cancelled_by_client | 2013-10-1 |
| 5 | 1 | 10 | 1 | completed | 2013-10-2 |
| 6 | 2 | 11 | 6 | completed | 2013-10-2 |
| 7 | 3 | 12 | 6 | completed | 2013-10-2 |
| 8 | 2 | 12 | 12 | completed | 2013-10-3 |
| 9 | 3 | 10 | 12 | completed | 2013-10-3 |
| 10 | 4 | 13 | 12 | cancelled_by_driver | 2013-10-3 |
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
DROP TABLE if EXISTS Trips;
CREATE TABLE Trips
(Id INT,
Client_Id INT,
Driver_Id INT,
City_Id INT,
Status VARCHAR(30),
Request_at DATE,
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');
SELECT * FROM Trips;
DROP TABLE if EXISTS Users ;
CREATE TABLE Users
(Users_Id INT,
Banned VARCHAR(30),
Role VARCHAR(30),
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 * FROM Users;
-- QUERY
SELECT TTT.Request_at,
-- 3.分别对符合条件的分子和分母求和，并保留2位小数
ROUND(SUM(CASE WHEN TTT.Status like 'cancelled%' AND TTT.Banned1 != 'YES' AND TTT.Banned2 != 'YES' THEN 1 ELSE 0 END)/SUM(CASE WHEN TTT.Banned1 != 'YES' AND TTT.Banned2 != 'YES' THEN 1 ELSE 0 END),2) AS "Cancellation Rate"
FROM
(
-- 2.获取 driver 用户 Banned 标识
SELECT TT.Request_at, TT.Status, TT.Banned AS Banned1, UU.Banned AS Banned2
FROM
(
-- 1.获取 cilent 用户 Banned 标识
SELECT T.Request_at, T.Driver_Id, T.Status, U.Banned, U.Role FROM Trips T
JOIN Users U
ON T.Client_Id = U.Users_Id AND U.Role = 'client'
WHERE T.Request_at BETWEEN '2013-10-01' AND '2020-10-03') TT
JOIN Users UU
ON TT.Driver_Id = UU.Users_Id AND UU.Role = 'driver') TTT
GROUP BY TTT.Request_at
