LeetCode 262_行程和用户
丹丹
 2019-09-08 00:00:00    448 Words  2 Mins  

题目描述

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’)。

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+----+-----------+-----------+---------+--------------------+----------+
| 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。

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+----------+--------+--------+
| 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。

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

解题思路

  1. 将最后的限制先做出来，并保留唯一识别ID
  2. 分组记数
  3. 保留两位应用format，但官网解答写错用了round才对

code

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# Write your MySQL query statement below
select
    Request_at as 'Day',
    round(sum(case when Status in ("cancelled_by_driver","cancelled_by_client") then 1 else 0 end) / count(*),2) as 'Cancellation Rate'

from Trips
where id in 
(select
    a.id as 'id'
from Trips a
left join Users b
    on a.Client_Id = b.Users_Id
left join Users c
    on a.Driver_Id = c.Users_Id
where b.Banned = "No" and c.Banned = "No" and Request_at >= "2013-10-01" and Request_at <= "2013-10-03")

group by Request_at

成绩

  • 用时：151 ms
  • 内存消耗：N/A

参考：

  1. 262. 行程和用户 - 力扣（LeetCode）