作业1：

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 **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。

+------------+-------------------+

| Day | Cancellation Rate |

+------------+-------------------+

| 2013-10-01 | 0.33 |

| 2013-10-02 | 0.00 |

| 2013-10-03 | 0.50 |

+------------+-------------------+

SELECT Request_at AS Day, ROUND(count(Status != 'completed' OR NULL)/count(*),2) AS "Cancellation Rate"

FROM Trips, users

WHERE Trips.Client_Id = users.Users_Id

AND Banned="no" and Request_at BETWEEN "2013-10-1" AND "2013-10-3"

GROUP BY Request_at;

作业2:

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 |

+------------+----------+--------+

select d.Name Department, e.Name Employee, e.Salary Salary

from Employee e inner join Department d

on e.department_id = d.Id

where (

select count(distinct c.Salary)

from Employee c

where e.Salary < c.Salary and e.department_id = c.department_id

) < 3

order by e.department_id, e.Salary desc;

作业3:

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 |

+-------+------+

SELECT score,(SELECT COUNT(score)+1

FROM score

WHERE score>s.score) AS r

FROM score AS s

ORDER BY score DESC;