Mysql练习生：新手上路项目十六——二十

连接Mysql

d：
cd D:\mysql-5.7.27-winx64\mysql-5.7.27-winx64\bin
net start mysql
mysql -u root -p

use DataWhaletwo;

CREATE TABLE score(
id int,
score DECIMAL(3,2)
);

INSERT INTO score
VALUES
(1,3.50),
(2,3.65),
(3,4.00),
(4,3.85),
(5,4.00),
(6,3.65);

mysql> select @rank_tmp:=0,@pre_score:=NULL;

SELECT id, score
 , CASE 
  WHEN @pre_score = score THEN @rank_tmp
  WHEN @pre_score := score THEN @rank_tmp := @rank_tmp + 1
 END AS 'rank'
FROM score, (
  SELECT @rank_tmp := 0, @pre_score := NULL
 ) tmp
ORDER BY score.score DESC;

项目十七：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

CREATE TABLE survey_log(
id int PRIMARY KEY,
uid int(11) ,
action varchar(20),
question_id int(11),
answer_id int(11),
q_num int(11),
timestamp varchar(20)
);

INSERT INTO survey_log VALUES
('1', '5', 'show', '285', null, '1', '123'),
('2', '5', 'answer', '285', '124124', '1', '124'),
('3', '5', 'show', '369', null, '2', '125'),
('4', '5', 'skip', '369', null, '2', '126');

SELECT sum(question_id) FROM survey_log WHERE action LIKE 'show';
SELECT SUM(CASE 
		WHEN action LIKE 'answer' THEN 1
		ELSE 0
	END)
FROM survey_log;

SELECT SUM(CASE 
  WHEN action LIKE 'answer' THEN 1 
  ELSE 0
 END) / SUM(CASE 
  WHEN action LIKE 'show' THEN 1
  ELSE 0
 END) AS rate, question_id
FROM survey_log
GROUP BY question_id
ORDER BY rate DESC;
SELECT question_id AS survey_log
FROM (
 SELECT SUM(CASE 
   WHEN action LIKE 'answer' THEN 1
   ELSE 0
  END) / SUM(CASE 
   WHEN action LIKE 'show' THEN 1
   ELSE 0
  END) AS rate, question_id
 FROM survey_log
 GROUP BY question_id
 ORDER BY rate DESC
 LIMIT 1
) tmp;

项目十八：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。

DELETE FROM Employee;

INSERT INTO Employee(Id,Name,Salary,DepartmentId) VALUES
('1','Joe','70000','1'),
('2','Henry','80000','2'),
('3','Sam','60000',''2'),
('4','Max','90000','1'),
('5',"Janet",69000,1),
('6',"Randy",85000,1);

SELECT D1.NAME AS DepartmentId, E1.NAME AS employee, E1.Salary
FROM employee E1, employee E2, department D1
WHERE (E1.DepartmentId = E2.DepartmentId
 AND E2.Salary >= E1.Salary
 AND E1.DepartmentId = D1.Id)
GROUP BY E1.Name
HAVING COUNT(DISTINCT E2.Salary) <= 3
ORDER BY D1.Name, E1.Salary DESC;

项目十九：平面上最近距离

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：

注意： 所有点的最大距离小于10000。

// An highlighted block
var foo = 'bar';
CREATE TABLE point_2d (
x int,
y int
);

INSERT INTO point_2d VALUES 
(-1, -1),(0, 0),(-1, -2);
SELECT MIN(distance) AS shortest
  FROM (
    SELECT POW(P1.x - p2.x, 2) + POW(p1.y - p2.y, 2) AS distance
    FROM point_2d p1
    INNER JOIN point_2d p2 ON !(p1.x = p2.x
    AND p1.y = p2.y) 
   ) as p
   

项目二十：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’)。

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。

// An highlighted block
var foo = 'bar';
DROP TABLE if EXISTS Trips;
DROP TABLE if EXISTS Users;

CREATE TABLE Users(
Users_Id INT UNIQUE,
Banned VARCHAR(5),
Role VARCHAR(10)		
);

CREATE TABLE Trips(
Id INT UNIQUE,
Client_Id INT,
Driver_Id INT,
City_Id INT,
Status VARCHAR(30),
Request_at Date,
CONSTRAINT FOREIGN KEY (Client_Id) REFERENCES Users(Users_Id),
CONSTRAINT FOREIGN KEY(Driver_Id) REFERENCES Users(Users_Id)
);

INSERT INTO Users
VALUES (1, 'No', 'client'),
 (2, 'Yes', 'client'),
 (3, 'No', 'client'),
 (4, 'No', 'client'),
 (10, 'No', 'driver'),
 (11, 'No', 'driver'),
 (12, 'No', 'driver'),
 (13, 'No', 'driver');

INSERT INTO Trips
VALUES (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, 11, 12, 'completed'
  , '2013-10-03'),
 (9, 3, 10, 12, 'completed'
  , '2013-10-03'),
 (10, 4, 13, 12, 'completed_by_driver'
  , '2013-10-03');

SELECT Users_Id AS userid
FROM Users
WHERE Banned = 'Yes'
AND Role = 'client';

WHERE Client_id NOT IN (  
	SELECT Users_Id AS userid
	FROM Users
	WHERE Banned = 'Yes'
		AND Role = 'client';

SELECT SUM(Status != 'completed') / COUNT(*) 
FROM trips
GROUP BY Request_at

ELECT Request_at AS Day
 , SUM(Status != 'completed') / COUNT(*) AS 'Cancellation Rate' 
FROM trips
WHERE Client_Id NOT IN (          
  SELECT Users_Id AS userid      
  FROM Users
  WHERE Banned = 'Yes'
   AND Role = 'client'
 )
 AND DATEDIFF(Request_at, '2013-10-01') >= 0  
 AND DATEDiFF(Request_at, '2013-10-01') <= 2
GROUP BY Request_at
;