Datawhale MySQL 训练营 Task6 实战项目

时间 2019-12-13

标签 datawhale mysql 训练营 task6 task 实战项目栏目 MySQL 繁體版

原文原文链接

做业

项目十：行程和用户（难度：困难）
Trips 表中存全部出租车的行程信息。每段行程有惟一键 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 表存全部用户。每一个用户有惟一键 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。sql

+----------+--------+--------+
| 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 语句应返回以下结果，取消率（Cancellation Rate）保留两位小数。函数

+------------+-------------------+
|     Day    | Cancellation Rate |
+------------+-------------------+
| 2013-10-01 |       0.33        |
| 2013-10-02 |       0.00        |
| 2013-10-03 |       0.50        |
+------------+-------------------+

解答：
建立表格code

CREATE TABLE IF NOT EXISTS Trips (
Id         INT, 
Client_Id  INT, 
Driver_Id  INT, 
City_Id    INT, 
Status     ENUM('completed', 'cancelled_by_driver', 'cancelled_by_client'), 
Request_at VARCHAR(50)
);

CREATE TABLE IF NOT EXISTS Users (
Users_Id INT, 
Banned   VARCHAR(50), 
Role     ENUM('client', 'driver', 'partner')
);


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');

查询blog

SELECT
    a.Request_at AS Day, ROUND(a.cnt/b.total,2) AS 'Cancellation Rate'
FROM
    (
    SELECT
        Request_at,
        COUNT( Id ) AS cnt 
    FROM
        trips 
    WHERE
        Client_Id IN ( SELECT Users_Id FROM users WHERE Banned = 'NO' AND Role = 'client' ) 
        AND `Status` = 'cancelled_by_driver' 
    GROUP BY
        Request_at 
    ) AS a,
    (
    SELECT
        Request_at,
        COUNT( Id ) AS total 
    FROM
        trips 
    WHERE
    Client_Id IN ( SELECT Users_Id FROM users WHERE Banned = 'NO' AND Role = 'client' ) 
    GROUP BY
    Request_at
    ) AS b;

而后我发现我作得不对，我只考虑了被司机取消得订单。。。尬住。
杨大大的答案：ip

SELECT
    Request_at AS DAY,
    ROUND( SUM( CASE WHEN `Status` LIKE 'cancelled%' THEN 1 ELSE 0 END ) / COUNT( * ), 2 ) AS cancelled_rate 
FROM
    trips
    LEFT JOIN ( SELECT Users_Id, Banned FROM users ) client ON trips.Client_Id = client.Users_Id
    LEFT JOIN ( SELECT Users_Id, Banned FROM users ) driver ON trips.Driver_Id = driver.Users_Id
    WHERE client.Banned = 'NO' AND driver.Banned = 'NO'
    GROUP BY Request_at
    ORDER BY Request_at;

项目十一：各部门前3高工资的员工（难度：中等）

将昨天employee表清空，从新插入如下数据（实际上是多插入5,6两行）：it

+----+-------+--------+--------------+
| 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，找出每一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：io

+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT         | Max      | 90000  |
| IT         | Randy    | 85000  |
| IT         | Joe      | 70000  |
| Sales      | Henry    | 80000  |
| Sales      | Sam      | 60000  |
+------------+----------+--------+

TRUNCATE TABLE employee;
INSERT INTO employee
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
    department.`Name` AS Department,
    a.`Name` AS `Name`,
    a.Salary AS Salary 
FROM
    employee AS a
    INNER JOIN department ON department.Id = a.DepartmentId 
WHERE
    (
    SELECT
        COUNT( DISTINCT Salary ) 
    FROM
        employee AS b /*辅助表技术*/
    WHERE
        b.DepartmentId = a.DepartmentId 
        AND b.Salary >= a.Salary 
    ) <= 3 
ORDER BY
    DepartmentId,
    Salary DESC;

任务十二：分数排名 - （难度：中等）
依然是昨天的分数表，实现排名功能，可是排名是非连续的，以下：

+-------+------+
| Score | Rank |
+-------+------+
| 4.00  | 1    |
| 4.00  | 1    |
| 3.85  | 3    |
| 3.65  | 4    |
| 3.65  | 4    |
| 3.50  | 6    |
+-------+------+

解答：
最简单的
思路直接就是使用 MySQL 8.x支持的窗口函数class

SELECT
    id,
    score,
    rank ( ) over ( ORDER BY score.score DESC ) AS 'rank' 
FROM
    score 
ORDER BY
    score DESC;

结果：
cli