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 |
±-----------±---------±-------+
CREATE TABLE Employee(
Id INT NOT NULL PRIMARY KEY,
Name VARCHAR(255) NOT NULL,
Salary INT,
DepartmentId INT
);
INSERT INTO Employee (Id,Name,Salary,DepartmentId)
VALUES (1,'Joe',70000,1),(2,'Henry',80000,2),(3,'Sam',60000,2),(4,'Max',90000,2);
CREATE TABLE Department(
Id INT NOT NULL,
Name VARCHAR(255)
);
INSERT INTO Department (Id,Name)
VALUES (1,'IT'),(2,'Sales');
SELECT Department.Name As Department, Employee.Name As Employee, Employee.Salary As Salary
FROM Department JOIN Employee
ON Department.Id = Employee.DepartmentId
WHERE Salary IN (SELECT MAX(Salary) FROM Employee GROUP BY DepartmentId);
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 NOT NULL PRIMARY KEY,
student VARCHAR(255)
);
INSERT INTO seat
VALUES
(1,'Abbot'),
(2,'Doris'),
(3,'Emerson'),
(4,'Green'),
(5,'Jeames');
SELECT
( CASE when MOD(id,2)=1 AND id!=(SELECT MAX(id) FROM seat ) then id+1
when MOD(ID,2)=0 then id-1
ELSE id
END) id ,student
FROM seat
ORDER BY id;
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 score(
Id INT,
Score DECIMAL(3,2)
);
INSERT INTO score
VALUES (1,3.50),(2,3.65),(3,4.00),(4,3.85),(5,4.00),(6,3.65);
SELECT Score,(
SELECT COUNT(DISTINCT Score)
FROM score
WHERE Score >= s.Score) AS 'Rank'
FROM score AS s
ORDER BY Score DESC;
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 |
±-----------±------------------+
CREATE TABLE Trips (
Id INT PRIMARY KEY,
Client_Id INT,
Driver_Id INT,
City_Id INT,
Status ENUM('completed', 'cancelled_by_driver', 'cancelled_by_client'),
Request_at VARCHAR(50)
);
CREATE TABLE Users (
Users_Id INT PRIMARY KEY,
Banned VARCHAR(50),
Role ENUM('client', 'driver', 'partner')
);
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at) VALUES ('1', '1', '10', '1', 'completed', '2013-10-01');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at) VALUES ('2', '2', '11', '1', 'cancelled_by_driver', '2013-10-01');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at) VALUES ('3', '3', '12', '6', 'completed', '2013-10-01');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at) VALUES ('4', '4', '13', '6', 'cancelled_by_client', '2013-10-01');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at) VALUES ('5', '1', '10', '1', 'completed', '2013-10-02');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at) VALUES ('6', '2', '11', '6', 'completed', '2013-10-02');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at) VALUES ('7', '3', '12', '6', 'completed', '2013-10-02');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at) VALUES ('8', '2', '12', '12', 'completed', '2013-10-03');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at) VALUES ('9', '3', '10', '12', 'completed', '2013-10-03');
INSERT INTO Trips (Id, Client_Id, Driver_Id, City_Id, Status, Request_at) VALUES ('10', '4', '13', '12', 'cancelled_by_driver', '2013-10-03');
INSERT INTO Users (Users_Id, Banned, Role) VALUES ('1', 'No', 'client');
INSERT INTO Users (Users_Id, Banned, Role) VALUES ('2', 'Yes', 'client');
INSERT INTO Users (Users_Id, Banned, Role) VALUES ('3', 'No', 'client');
INSERT INTO Users (Users_Id, Banned, Role) VALUES ('4', 'No', 'client');
INSERT INTO Users (Users_Id, Banned, Role) VALUES ('10', 'No', 'driver');
INSERT INTO Users (Users_Id, Banned, Role) VALUES ('11', 'No', 'driver');
INSERT INTO Users (Users_Id, Banned, Role) VALUES ('12', 'No', 'driver');
INSERT INTO Users (Users_Id, Banned, Role) VALUES ('13', 'No', 'driver');
ALTER TABLE Trips
ADD FOREIGN KEY(Client_Id)
REFERENCES Users(Users_Id);
ALTER TABLE Trips
ADD FOREIGN KEY(Driver_Id)
REFERENCES Users(Users_Id);
SELECT Request_at as Day,
ROUND(SUM(CASE when Status LIKE 'Cancelled%' then 1 else 0 END)/COUNT(*),2) as 'Cancellation Rate'
FROM Trips JOIN Users
ON Users.Users_Id = Trips.Client_Id AND Users.Banned = 'No'
GROUP BY Request_at;
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。
TRUNCATE TABLE employee;
INSERT INTO Employee
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 as Department,e.Name as Employee,e.Salary as Salary
FROM Employee e
JOIN Department d ON e.DepartmentId = d.Id
WHERE Salary IN(
SELECT Salary
FROM Employee
WHERE (SELECT COUNT(DISTINCT Salary)
FROM Employee
WHERE Salary >= e.Salary AND DepartmentId=d.id) <=3)
ORDER BY DepartmentId,Salary DESC;
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)
FROM score
WHERE Score > s.Score) +1) as 'Rank'
FROM score as s
ORDER BY Score DESC;