sql练习（行程和路线）

最新推荐文章于 2024-07-21 21:09:39 发布

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最新推荐文章于 2024-07-21 21:09:39 发布

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这段SQL查询旨在确定在2013年10月1日至3日期间，未被禁止的用户（乘客和司机）的行程取消率。取消率是通过计算被取消的行程（由司机或客户取消）数量除以总行程数量得出的，结果保留两位小数。查询结果展示每日的取消率。

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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

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              |
+------------+-------------------+

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/trips-and-users

我的解答（当然在实际运用中要么用存储过程，要么创建临时表，这样可以加快速度）：

select d.Request_at as `Day`,ROUND((
    select count(ts.Id) from Trips as ts where ts.Request_at=t.Request_at and (ts.Status='cancelled_by_driver' or ts.Status='cancelled_by_client') and
    exists(
        select 1 from Users where
        Banned='No' and
        (Users_Id=ts.Client_Id)
        and (Role='client')
    )
    and
    exists(
        select 1 from Users where
        Banned='No' and
        (Users_Id=ts.Driver_Id)
        and (Role='driver')
    )
)/count(t.Id),2) as `Cancellation Rate` from (
    select tt.* from Trips as tt where
    exists(
        select 1 from Users where
        Banned='No' and
        (Users_Id=tt.Client_Id)
        and (Role='client')
    )
    and
    exists(
        select 1 from Users where
        Banned='No' and
        (Users_Id=tt.Driver_Id)
        and (Role='driver')
    )
) as t right join (
    select distinct(Request_at) from Trips
) as d 
on d.Request_at=t.Request_at
where (d.Request_at Between cast('2013-10-01' as DATE) and cast('2013-10-03' as DATE))
group by t.Request_at
having count(t.Id)>0
order by t.Request_at