MySQL行程和用户

逃离我推掉我的手 2023-10-06 18:51 7阅读 0赞

#### MySQL行程和用户 ####

*  SQL架构
 *  题目描述
 *  题解
 *   *  解法一
     *  解法二
     *  解题思路

## SQL架构 ##

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'));
    
    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');
    
    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');

## 题目描述 ##

表：Trips

+-------------+----------+
    | Column Name | Type     |
    +-------------+----------+
    | Id          | int      |
    | Client_Id   | int      |
    | Driver_Id   | int      |
    | City_Id     | int      |
    | Status      | enum     |
    | Request_at  | date     |     
    +-------------+----------+

Id 是这张表的主键。

这张表中存所有出租车的行程信息。每段行程有唯一 Id ，其中 Client\_Id 和 Driver\_Id 是 Users 表中 Users\_Id 的外键。

Status 是一个表示行程状态的枚举类型，枚举成员为(‘completed’, ‘cancelled\_by\_driver’, ‘cancelled\_by\_client’) 。

表：Users

Users\_Id 是这张表的主键。

这张表中存所有用户，每个用户都有一个唯一的 Users\_Id ，Role 是一个表示用户身份的枚举类型，枚举成员为 (‘client’, ‘driver’, ‘partner’) 。

Banned 是一个表示用户是否被禁止的枚举类型，枚举成员为 (‘Yes’, ‘No’) 。

> 写一段 SQL 语句查出 “2013-10-01” 至 “2013-10-03” 期间非禁止用户（乘客和司机都必须未被禁止）的取消率。非禁止用户即 Banned 为 No 的用户，禁止用户即 Banned 为 Yes 的用户。

取消率 的计算方式如下：(被司机或乘客取消的非禁止用户生成的订单数量) / (非禁止用户生成的订单总数)。

返回结果表中的数据可以按任意顺序组织。其中取消率 Cancellation Rate 需要四舍五入保留 两位小数 。

查询结果格式如下例所示：

Trips 表：
    +----+-----------+-----------+---------+---------------------+------------+
    | 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：
    +----------+--------+--------+
    | 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 |
    +----------+--------+--------+
    
    Result 表：
    +------------+-------------------+
    | Day        | Cancellation Rate |
    +------------+-------------------+
    | 2013-10-01 | 0.33              |
    | 2013-10-02 | 0.00              |
    | 2013-10-03 | 0.50              |
    +------------+-------------------+
    
    2013-10-01：
      - 共有 4 条请求，其中 2 条取消。
      - 然而，Id=2 的请求是由禁止用户（User_Id=2）发出的，所以计算时应当忽略它。
      - 因此，总共有 3 条非禁止请求参与计算，其中 1 条取消。
      - 取消率为 (1 / 3) = 0.33
    2013-10-02：
      - 共有 3 条请求，其中 0 条取消。
      - 然而，Id=6 的请求是由禁止用户发出的，所以计算时应当忽略它。
      - 因此，总共有 2 条非禁止请求参与计算，其中 0 条取消。
      - 取消率为 (0 / 2) = 0.00
    2013-10-03：
      - 共有 3 条请求，其中 1 条取消。
      - 然而，Id=8 的请求是由禁止用户发出的，所以计算时应当忽略它。
      - 因此，总共有 2 条非禁止请求参与计算，其中 1 条取消。
      - 取消率为 (1 / 2) = 0.50

## 题解 ##

统计每天非禁止用户的取消率，需要知道非禁止用户有哪些，总行程数，取消的行程数。

### 解法一 ###

首先确定被禁止用户的行程记录，再剔除这些行程记录。

行程表中，字段 client\_id 和 driver\_id，都与用户表中的 users\_id 关联。因此只要 client\_id 和 driver\_id 中有一个被禁止了，此条行程记录要被剔除。

先说一种错误的找出没被禁止用户行程记录的方法。此方法很有迷惑性。

思路：

if (client_id = users_id 或 driver_id = users_id) 且 users_id没有被禁止
    {
        此条记录没被禁止。
    }

SELECT *
    FROM Trips AS T JOIN Users AS U 
    ON (T.client_id = U.users_id  OR T.driver_id = U.users_id )  AND U.banned ='No';

乍一看，思路是对。其实是错误的。因为，我们不知觉得肯定了一个假设—— client\_id 与 driver\_id 是相同的。只有当两者相同时，才能用此条件排除被禁止用户的行程记录。

错误的结果：

+------+-----------+-----------+---------+---------------------+------------+----------+--------+--------+
    | Id   | Client_Id | Driver_Id | City_Id | Status              | Request_at | Users_Id | Banned | Role   |
    +------+-----------+-----------+---------+---------------------+------------+----------+--------+--------+
    |    1 |         1 |        10 |       1 | completed           | 2013-10-01 |        1 | No     | client |
    |    1 |         1 |        10 |       1 | completed           | 2013-10-01 |       10 | No     | driver |
    |    2 |         2 |        11 |       1 | cancelled_by_driver | 2013-10-01 |       11 | No     | driver |
    |    3 |         3 |        12 |       6 | completed           | 2013-10-01 |        3 | No     | client |
    |    3 |         3 |        12 |       6 | completed           | 2013-10-01 |       12 | No     | driver |
    |    4 |         4 |        13 |       6 | cancelled_by_client | 2013-10-01 |        4 | No     | client |
    |    4 |         4 |        13 |       6 | cancelled_by_client | 2013-10-01 |       13 | No     | driver |
    |    5 |         1 |        10 |       1 | completed           | 2013-10-02 |        1 | No     | client |
    |    5 |         1 |        10 |       1 | completed           | 2013-10-02 |       10 | No     | driver |
    |    6 |         2 |        11 |       6 | completed           | 2013-10-02 |       11 | No     | driver |
    |    7 |         3 |        12 |       6 | completed           | 2013-10-02 |        3 | No     | client |
    |    7 |         3 |        12 |       6 | completed           | 2013-10-02 |       12 | No     | driver |
    |    8 |         2 |        12 |      12 | completed           | 2013-10-03 |       12 | No     | driver |
    |    9 |         3 |        10 |      12 | completed           | 2013-10-03 |        3 | No     | client |
    |    9 |         3 |        10 |      12 | completed           | 2013-10-03 |       10 | No     | driver |
    |   10 |         4 |        13 |      12 | cancelled_by_driver | 2013-10-03 |        4 | No     | client |
    |   10 |         4 |        13 |      12 | cancelled_by_driver | 2013-10-03 |       13 | No     | driver |
    +------+-----------+-----------+---------+---------------------+------------+----------+--------+--------+
    17 rows in set (0.00 sec)

结果中，被禁止的 users\_id = 2，其行程记录没被剔除掉。

明显， client\_id 与 driver\_id 不一定相同 。

正确的做法是对 client\_id 和 driver\_id 各自关联的 users\_id，同时检测是否被禁止。

if (client_id = users_id_1 且 users_id_1没被禁止 并且 client_id = users_id_2 且 users_id_2没被禁止){
        此条记录没被禁止。
    }

SELECT *
    FROM Trips AS T
    JOIN Users AS U1 ON (T.client_id = U1.users_id AND U1.banned ='No')
    JOIN Users AS U2 ON (T.driver_id = U2.users_id AND U2.banned ='No');
    
    +------+-----------+-----------+---------+---------------------+------------+----------+--------+--------+----------+--------+--------+
    | Id   | Client_Id | Driver_Id | City_Id | Status              | Request_at | Users_Id | Banned | Role   | Users_Id | Banned | Role   |
    +------+-----------+-----------+---------+---------------------+------------+----------+--------+--------+----------+--------+--------+
    |    1 |         1 |        10 |       1 | completed           | 2013-10-01 |        1 | No     | client |       10 | No     | driver |
    |    3 |         3 |        12 |       6 | completed           | 2013-10-01 |        3 | No     | client |       12 | No     | driver |
    |    4 |         4 |        13 |       6 | cancelled_by_client | 2013-10-01 |        4 | No     | client |       13 | No     | driver |
    |    5 |         1 |        10 |       1 | completed           | 2013-10-02 |        1 | No     | client |       10 | No     | driver |
    |    7 |         3 |        12 |       6 | completed           | 2013-10-02 |        3 | No     | client |       12 | No     | driver |
    |    9 |         3 |        10 |      12 | completed           | 2013-10-03 |        3 | No     | client |       10 | No     | driver |
    |   10 |         4 |        13 |      12 | cancelled_by_driver | 2013-10-03 |        4 | No     | client |       13 | No     | driver |
    +------+-----------+-----------+---------+---------------------+------------+----------+--------+--------+----------+--------+--------+
    7 rows in set (0.00 sec)

在此基础上，按日期分组，统计每组的 总行程数，取消的行程数 。

每组的总行程数：COUNT(T.STATUS)。

每组取消的行程数：

SUM(
    	IF(T.STATUS = 'completed',0,1)
    )

取消率 = 每组的取消的行程数 / 每组的总行程数

完整逻辑为:

SELECT T.request_at AS `Day`, 
    	ROUND(
    			SUM(
    				IF(T.STATUS = 'completed',0,1)
    			)
    			/ 
    			COUNT(T.STATUS),
    			2
    	) AS `Cancellation Rate`
    FROM Trips AS T
    JOIN Users AS U1 ON (T.client_id = U1.users_id AND U1.banned ='No')
    JOIN Users AS U2 ON (T.driver_id = U2.users_id AND U2.banned ='No')
    WHERE T.request_at BETWEEN '2013-10-01' AND '2013-10-03'
    GROUP BY T.request_at;
    
    +------------+-------------------+
    | Day        | Cancellation Rate |
    +------------+-------------------+
    | 2013-10-01 |              0.33 |
    | 2013-10-02 |              0.00 |
    | 2013-10-03 |              0.50 |
    +------------+-------------------+
    3 rows in set (0.01 sec)

其中 SUM 求和函数，COUNT 计数函数，ROUND 四舍五入函数。

### 解法二 ###

与解法二思路相同。找出被禁止的用户后，不再连接行程表和用户表，直接从行程表中排除掉被被禁止用户的行程记录。

被禁止的用户用子查询：

SELECT T.request_at AS `Day`, 
    	ROUND(
    			SUM(
    				IF(T.STATUS = 'completed',0,1)
    			)
    			/ 
    			COUNT(T.STATUS),
    			2
    	) AS `Cancellation Rate`
    FROM Trips AS T
    WHERE 
    T.Client_Id NOT IN (
    	SELECT users_id
    	FROM Users
    	WHERE banned = 'Yes'
    )
    AND
    T.Driver_Id NOT IN (
    	SELECT users_id
    	FROM Users
    	WHERE banned = 'Yes'
    )
    AND T.request_at BETWEEN '2013-10-01' AND '2013-10-03'
    GROUP BY T.request_at;
    
    +------------+-------------------+
    | Day        | Cancellation Rate |
    +------------+-------------------+
    | 2013-10-01 |              0.33 |
    | 2013-10-02 |              0.00 |
    | 2013-10-03 |              0.50 |
    +------------+-------------------+
    3 rows in set (0.01 sec)

### 解题思路 ###

trips与2个users表连结，分别用dirver\_id和client\_id连结，然后2个user表用banned='No’确保都是未被禁止的用户。

然后针对t.Request\_at分组。

在查询结果里面，用case语句对完成的订单进行计数，除以所有订单数得到率

代码

SELECT
    	t.Request_at AS Day,
    	round( sum( CASE t.`Status` WHEN 'completed' THEN 0 ELSE 1 END ) / count( * ), 2 ) AS `Cancellation Rate` 
    FROM
    	Trips t,
    	Users d,
    	Users c 
    WHERE
    	t.Driver_Id = d.Users_Id 
    	AND t.Client_Id = c.Users_Id 
    	AND d.Banned = 'No' 
    	AND c.Banned = 'No' 
    	AND t.Request_at BETWEEN STR_TO_DATE('2013-10-01','%Y-%m-%d') and STR_TO_DATE('2013-10-03','%Y-%m-%d') 
    GROUP BY
    	t.Request_at;