專案十：行程和使用者（難度：困難）
trips 表中存所有計程車的行程資訊。每段行程有唯一鍵 id，client_id 和 driver_id 是 users 表中 users_id 的外來鍵。status 是列舉型別，列舉成員為 (『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
建表：
create table trips(
id int not null primary key,
client_id int,
driver_id int,
city_id int,
status enum("completed", "cancelled_by_driver", "cancelled_by_client"),
request_at date,
foreign key(client_id) references users(users_id),
foreign key(driver_id) references users(users_id)
);
insert into trips values
(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");
建表：
create table users(
users_id int primary key,
banned enum("yes", "no"),
role enum("client", "driver", "role")
);
insert into users values
(1 ,"no","client"),
(2 ,"yes","client"),
(3 ,"no","client"),
(4 ,"no","client"),
(10 ,"no","driver"),
(11 ,"no","driver"),
(12 ,"no","driver"),
(13 ,"no","driver");
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;
專案十一：各部門前3高工資的員工（難度：中等）
將專案7中的employee表清空，重新插入以下資料（其實是多插入5,6兩行）： | 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 |
建表：
create table employee2(
id int not null,
name char(10) not null,
salary long,
departmentid int
);
insert into employee2 values
(1,"joe",70000,1),
(2,"henry",80000,2),
(3,"sam",60000,2),
(4,"max",90000,1),
(5,"janet",69000,1),
(6,"randy",85000,1);
select d.name department, e.name employee2, e.salary salary
from employee2 e inner join department d
on e.departmentid = d.id
where (
select count(distinct c.salary)
from employee2 c
where e.salary < c.salary and e.departmentid = c.departmentid
) < 3
order by e.departmentid, e.salary desc;
專案十二 分數排名 - （難度：中等）
依然是昨天的分數表，實現排名功能，但是排名是非連續的，如下： ±------±-----+ | score | rank | ±------±-----+ | 4.00 | 1 | | 4.00 | 1 | | 3.85 | 3 | | 3.65 | 4 | | 3.65 | 4 | | 3.50 | 6 | ±------±-----
select score as score,(
select count(score)+1
from scores s1
where s1.score>s2.score
) as "rank"
from scores s2
order by score desc;
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