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项目十六

项目十六 分数排名 （难度：中等）

依然是昨天的分数表，实现排名功能，但是排名需要是非连续的，如下： +-------+------+ | Score | Rank | +-------+------+ | 4.00 | 1 | | 4.00 | 1 | | 3.85 | 3 | | 3.65 | 4 | | 3.65 | 4 | | 3.50 | 6 | +-------+------

mysql> select score,
-> (select count(score)+1 from scores s1 where s1.score>s2.score)as rank
-> from scores s2
-> order by score desc;
+-------+------+
| score | rank |
+-------+------+
|  4.00 |1 |
|  4.00 |1 |
|  3.85 |3 |
|  3.65 |4 |
|  3.65 |4 |
|  3.50 |6 |
+-------+------+
6 rows in set (0.00 sec)

项目十七：查询回答率最高的问题 （难度：中等） 求出survey_log表中回答率最高的问题，表格的字段有：uid, action, question_id, answer_id, q_num, timestamp。 uid是用户id；action的值为：“show”， “answer”， “skip”；当action是"answer"时，answer_id不为空，相反，当action是"show"和"skip"时为空（null）；q_num是问题的数字序号。 写一条sql语句找出回答率最高的问题。 举例： 输入

12345678 +------+-----------+--------------+------------+-----------+------------+| uid | action | question_id | answer_id | q_num | timestamp |+------+-----------+--------------+------------+-----------+------------+| 5 | show | 285 | null | 1 | 123 || 5 | answer | 285 | 124124 | 1 | 124 || 5 | show | 369 | null | 2 | 125 || 5 | skip | 369 | null | 2 | 126 |+------+-----------+--------------+------------+-----------+------------+
输出
12345 +-------------+| survey_log |+-------------+| 285 |+-------------+
---- ----
说明
问题285的回答率为1/1，然而问题369的回答率是0/1，所以输出是285。
注意： 最高回答率的意思是：同一个问题出现的次数中回答的比例。 
select question_id as survey_log
from(
select (sum(case when `action` like 'answer' then 1 else 0 end) / sum(case when `action` like 'show' then 1 else 0 end)) as rate,question_id
from survey_log
group by question_id
order by rate desc
limit 1) x

项目十八：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。

select * from (
select b.Department,a.name,a.salary
row_number() over(partition by b.Department order by a.salary desc) rn
 from Employee a 
left join Department b on a.DepartmentId  = b.Id
) temp
where temp.rn<4 
order by temp.rn

项目十九：平面上最近距离

point_2d 表包含一个平面内一些点（超过两个）的坐标值（x，y）。 写一条查询语句求出这些点中的最短距离并保留2位小数。

x y
-1 -1
0 0
-1 -2

最短距离是1，从点（-1，-1）到点（-1，2）。所以输出结果为：

shortest
1.00

注意： 所有点的最大距离小于10000。

selectround(SQRT(min(power((p1.x-p2.x),2)+power((p1.y-p2.y),2))),2)"shortest"

frompoint_2d p1joinpoint_2d p2

onp1.x!=p2.xorp1.y!=p2.y;

项目二十：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
t.request_at as day
,round(count(distinct if(u1.banned = 'no' and u2.banned ='no' and t.status!='completed',t.id,null))/count(distinct if(u1.banned = 'no' and u2.banned ='no',t.id,null)),2) as "cancellation rate"
from trips t
left join users u1
on t.client_id=u1.users_id
left join users u2
on t.driver_id=u2.users_id
where t.request_at between '2013-10-01' and '2013-10-03'
group by 
t.request_at