LeetCode（数据库）- 行程和用户
Posted Lux_Sun
tags:
篇首语：本文由小常识网(cha138.com)小编为大家整理，主要介绍了LeetCode（数据库）- 行程和用户相关的知识，希望对你有一定的参考价值。
题目链接：点击打开链接
题目大意：略。
解题思路：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。
}SQL 代码
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 |
+------+-----------+-----------+---------+---------------------+------------+----------+--------+--------+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。
}SQL 代码：
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')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)。
每组的取消的行程数：
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其中 SUM 求和函数，COUNT 计数函数，ROUND 四舍五入函数。
- 解法二
思路与解法一相同。而采用不同的方法排除掉被禁止用户的行程记录。想到排除，就联想到集合差。
client_id 和 driver_id 的全部为集合 U。被禁止的 users_id 集合为 A。
U 减去 A 的结果为没被禁止的用户。
(
SELECT users_id
FROM users
WHERE banned = 'Yes'
) AS A好了，先演示一个错误的解法：
行程表连接表 A，排除掉被被禁止的行程。
SELECT *
FROM trips AS T,
(
SELECT users_id
FROM users
WHERE banned = 'Yes'
) AS A
WHERE (T.Client_Id != A.users_id AND T.Driver_Id != A.users_id)剩下的逻辑与解法一后部分相同，完善后的逻辑为：
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,
(
SELECT users_id
FROM users
WHERE banned = 'Yes'
) AS A
WHERE (T.Client_Id != A.users_id AND T.Driver_Id != A.users_id) AND T.request_at BETWEEN '2013-10-01' AND '2013-10-03'
GROUP BY T.request_at很可惜，当表 A 为空时，此方法的结果是空表。但是表 A 为空，可能是有用户但是没有被禁止的用户。因此方法是错误的。
正确的解法是：行程表 left join 表 A 两次，A.users_id 都为 NULL 的行都是没被排除的行。
SELECT *
FROM trips AS T LEFT JOIN
(
SELECT users_id
FROM users
WHERE banned = 'Yes'
) AS A ON (T.Client_Id = A.users_id)
LEFT JOIN (
SELECT users_id
FROM users
WHERE banned = 'Yes'
) AS A1
ON (T.Driver_Id = A1.users_id)
WHERE A.users_id IS NULL AND A1.users_id IS NULL补上其它部分的逻辑为：
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 LEFT JOIN
(
SELECT users_id
FROM users
WHERE banned = 'Yes'
) AS A ON (T.Client_Id = A.users_id)
LEFT JOIN (
SELECT users_id
FROM users
WHERE banned = 'Yes'
) AS A1
ON (T.Driver_Id = A1.users_id)
WHERE A.users_id IS NULL AND A1.users_id IS NULL AND T.request_at BETWEEN '2013-10-01' AND '2013-10-03'
GROUP BY T.request_at- 解法三
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 users_id
FROM users
WHERE banned = 'Yes'
)行程表中 client_id 和 driver_id 都在此子查询结果中的行要剔除掉。
SELECT *
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'
)补上其它部分：
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以上是关于LeetCode（数据库）- 行程和用户的主要内容，如果未能解决你的问题，请参考以下文章