行程和用户(难度:困难)

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’)。

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

+----------+--------+--------+ 
| 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。

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

各部门前3高工资的员工(难度:中等)

将项目7中的employee表清空,重新插入以下数据(其实是多插入5,6两行):

+----+-------+--------+-------------+ 
| Id | Name  | Salary | DepartmentId| 
+----+-------+--------+-------------+ 
| 1  | Joe   | 70000  |      1      | 
| 2  | Henry | 80000  |      2      | 
| 3  | Sam   | 60000  |      2      | 
| 4  | Max   | 90000  |      1      |  
| 5  | Janet | 69000  |      1      | 
| 6  | Randy | 85000  |      1      | 
+----+-------+--------+-------------+ 

编写一个 SQL 查询,找出每个部门工资前三高的员工。例如,根据上述给定的表格,查询结果应返回:

+------------+----------+--------+ 
| Department | Employee | Salary | 
+------------+----------+--------+ 
|     IT     |   Max    | 90000  | 
|     IT     |   Randy  | 85000  | 
|     IT     |   Joe    | 70000  | 
|    Sales   |   Henry  | 80000  | 
|    Sales   |   Sam    | 60000  | 
+------------+----------+--------+

此外,请考虑实现各部门前N高工资的员工功能。
MySQL任务四


分数排名 - (难度:中等)

依然是昨天的分数表,实现排名功能,但是排名是非连续的,如下:

+-------+------+ 
| Score | Rank | 
+-------+------+ 
| 4.00  |   1  | 
| 4.00  |   1  | 
| 3.85  |   3  | 
| 3.65  |   4  | 
| 3.65  |   4  | 
| 3.50  |   6  | 
+-------+------+

MySQL任务四

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