[LeetCode] Trips and Users 旅行和用户
The Trips table holds all taxi trips. Each trip has a unique Id, while Client_Id and Driver_Id are both foreign keys to the Users_Id at the Users table. Status is an ENUM type of (‘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| +----+-----------+-----------+---------+--------------------+----------+
The Users table holds all users. Each user has an unique Users_Id, and Role is an ENUM type of (‘client’, ‘driver’, ‘partner’).
+----------+--------+--------+ | 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 | +----------+--------+--------+
Write a SQL query to find the cancellation rate of requests made by unbanned clients between Oct 1, 2013 and Oct 3, 2013. For the above tables, your SQL query should return the following rows with the cancellation rate being rounded to two decimal places.
+------------+-------------------+ | Day | Cancellation Rate | +------------+-------------------+ | 2013-10-01 | 0.33 | | 2013-10-02 | 0.00 | | 2013-10-03 | 0.50 | +------------+-------------------+
这道题给了我们一个Trips表里面有一些Id和状态，还有请求时间，然后还有一个Users表，里面有顾客和司机的信息，然后有该顾客和司机有没有被Ban的信息，让我们返回一个结果看某个时间段内由没有被ban的顾客提出的取消率是多少，其实题目没有说清楚顾客到底包不包括司机，其实是包括的，由司机提出的取消请求也应计算进去，我们用Case When ... Then ... Else ... End关键字来做，我们用cancelled%来表示开头是cancelled的所有项，这样就包括了driver和client，然后分母是所有项，限制条件里限定了时间段，然后是没有被ban的，由于结果需要保留两位小数，所以我们用Round关键字且给定参数2即可，参见代码如下：
解法一：
SELECT t.Request_at Day, ROUND(SUM(CASE WHEN t.Status LIKE 'cancelled%' THEN 1 ELSE 0 END)/COUNT(*), 2) 'Cancellation Rate' FROM Trips t JOIN Users u ON t.Client_Id = u.Users_Id AND u.Banned = 'No' WHERE t.Request_at BETWEEN '2013-10-01' AND '2013-10-03' GROUP BY t.Request_at;
上面的Case When ... Then ... Else ... End关键字也可以用If关键字来替换，实现的效果一样：
解法二：
SELECT Request_at Day, ROUND(COUNT(IF(Status != 'completed', TRUE, NULL)) / COUNT(*), 2) 'Cancellation Rate' FROM Trips WHERE (Request_at BETWEEN '2013-10-01' AND '2013-10-03') AND Client_Id IN (SELECT Users_Id FROM Users WHERE Banned = 'No') GROUP BY Request_at;
参考资料：
https://leetcode.com/discuss/52594/sharing-my-solution
https://leetcode.com/discuss/96599/solution-without-join
