作业#
项目七: 各部门工资最高的员工(难度:中等)
创建Employee 表,包含所有员工信息,每个员工有其对应的 Id, salary 和 department Id。
+----+-------+--------+--------------+
| Id | Name | Salary | DepartmentId |
+----+-------+--------+--------------+
| 1 | Joe | 70000 | 1 |
| 2 | Henry | 80000 | 2 |
| 3 | Sam | 60000 | 2 |
| 4 | Max | 90000 | 1 |
+----+-------+--------+--------------+
创建Department 表,包含公司所有部门的信息。
+----+----------+
| Id | Name |
+----+----------+
| 1 | IT |
| 2 | Sales |
+----+----------+
编写一个 SQL 查询,找出每个部门工资最高的员工。例如,根据上述给定的表格,Max 在 IT 部门有最高工资,Henry 在 Sales 部门有最高工资。
+------------+----------+--------+
| Department | Employee | Salary |
+------------+----------+--------+
| IT | Max | 90000 |
| Sales | Henry | 80000 |
+------------+----------+--------+
- 创建表:
DROP TABLE IF EXISTS `employee`;
CREATE TABLE `employee` (
`Id` int(11) NOT NULL,
`Name` varchar(255) NOT NULL,
`Salary` int(11) NOT NULL,
`DepartmentId` int(11) NOT NULL,
PRIMARY KEY (`Id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
DROP TABLE IF EXISTS `department`;
CREATE TABLE `department` (
`Id` int(11) NOT NULL,
`Name` varchar(255) NOT NULL,
PRIMARY KEY (`Id`)
)ENGINE=InnoDB DEFAULT CHARSET=utf8;
INSERT INTO Employee VALUE('1','Joe','70000','1');
INSERT INTO Employee VALUE('2','Henry','80000','2');
INSERT INTO Employee VALUE('3','Sam','60000','2');
INSERT INTO Employee VALUE('4','Max','90000','1');
INSERT INTO Department VALUE('1','IT');
INSERT INTO Department VALUE('2','Sales');
SELECT
d.Name AS Department,
e.Name AS Employee,
e.Salary AS Salary
FROM
Employee AS e,
Department AS d
WHERE e.DepartmentId = d.Id AND
e.Salary >= (SELECT MAX(Salary) FROM Employee WHERE DepartmentId=d.Id)
ORDER BY Salary DESC;
项目八: 换座位(难度:中等)
小美是一所中学的信息科技老师,她有一张 seat 座位表,平时用来储存学生名字和与他们相对应的座位 id。
其中纵列的 id 是连续递增的
小美想改变相邻俩学生的座位。
你能不能帮她写一个 SQL query 来输出小美想要的结果呢?
请创建如下所示seat表:
示例:
+---------+---------+
| id | student |
+---------+---------+
| 1 | Abbot |
| 2 | Doris |
| 3 | Emerson |
| 4 | Green |
| 5 | Jeames |
+---------+---------+
假如数据输入的是上表,则输出结果如下:
+---------+---------+
| id | student |
+---------+---------+
| 1 | Doris |
| 2 | Abbot |
| 3 | Green |
| 4 | Emerson |
| 5 | Jeames |
+---------+---------+
注意:
如果学生人数是奇数,则不需要改变最后一个同学的座位。
CREATE TABLE seat(
id INT ,
student VARCHAR(20)
)
#插入数据
INSERT INTO seat VALUE('1','Abbot');
INSERT INTO seat VALUE('2','Doris');
INSERT INTO seat VALUE('3','Emerson');
INSERT INTO seat VALUE('4','Green');
INSERT INTO seat VALUE('5','Jeames');
SELECT *
FROM (
-- 遇到偶数往上移一个位置
SELECT id-1 AS id,student
FROM seat
WHERE id%2=0
UNION
-- 遇到奇数往下移一个位置
SELECT id+1 AS id,student
FROM seat
WHERE id%2=1 AND (id+1)<=(SELECT COUNT(*) FROM seat)
UNION
-- 处理最后一个位置,这里只考虑奇数情况,保持不变(偶数已经在第一步里处理了)
SELECT id AS id,student
FROM seat
WHERE id%2=1 AND(id+1)>(SELECT COUNT(*) FROM seat)
) AS c1
ORDER BY id ASC;
项目九: 分数排名(难度:中等)
编写一个 SQL 查询来实现分数排名。如果两个分数相同,则两个分数排名(Rank)相同。请注意,平分后的下一个名次应该是下一个连续的整数值。换句话说,名次之间不应该有“间隔”。
创建以下score表:
+----+-------+
| Id | Score |
+----+-------+
| 1 | 3.50 |
| 2 | 3.65 |
| 3 | 4.00 |
| 4 | 3.85 |
| 5 | 4.00 |
| 6 | 3.65 |
+----+-------+
例如,根据上述给定的 Scores 表,你的查询应该返回(按分数从高到低排列):
+-------+------+
| Score | Rank |
+-------+------+
| 4.00 | 1 |
| 4.00 | 1 |
| 3.85 | 2 |
| 3.65 | 3 |
| 3.65 | 3 |
| 3.50 | 4 |
+-------+------+
CREATE TABLE scores(
ID INT,
Score FLOAT);
INSERT INTO scores VALUE('1','3.50');
INSERT INTO scores VALUE('2','3.65');
INSERT INTO scores VALUE('3','4.00');
INSERT INTO scores VALUE('4','3.85');
INSERT INTO scores VALUE('5','4.00');
INSERT INTO scores VALUE('6','3.65');
SELECT score,(SELECT COUNT(DISTINCT score)
FROM scores
WHERE score>=s.score) AS ran_k
FROM scores AS s
ORDER BY score DESC;
4.2 MySQL 实战 - 复杂项目
作业#
项目十: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 |
+------------+-------------------+
- 创建表
CREATE TABLE trips(
Id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
Client_Id INT NOT NULL,
Driver_Id INT NOT NULL,
City_Id INT NOT NULL,
Status ENUM(‘completed’, ‘cancelled_by_driver’, ‘cancelled_by_client’) NOT NULL,
Request_at DATE DEFAULT NULL);
CREATE TABLE Users(
Users_Id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
Banned VARCHAR(10) NOT NULL,
Role ENUM(‘client’, ‘driver’, ‘partnet’) NOT NULL);
- 插入数据
NSERT 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’);
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’);
- 查询表:
SELECT t.Request_at AS ‘Day’,
ROUND((SUM(CASE WHEN t.Status LIKE ‘cancelled%’ THEN 1 ELSE 0 END))/COUNT(*),2) AS ‘Cancellation Rate’
FROM Trips AS t INNER JOIN Users AS u
ON u.Users_Id = t.Client_Id AND u.Banned = ‘No’
GROUP BY t.Request_at ;
项目十一: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。
项目十二 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 |
+-------+------