Mysql数据库两表连接进行各种操作

一，创建两个表emp和dept，并给它们插入数据

1.创建表emp

create table dept (dept1 int ,dept_name varchar(11)) charset=utf8;

2.创建表dept
create table emp (sid int ,name varchar(11),age int,worktime_start date,incoming int,dept2 int) charset=utf8;
3.为dept表输入数据

insert into dept values
(101,'财务'),
(102,'销售'),
(103,'IT技术'),
(104,'行政');

4.为emp表输入数据

insert into emp values
(1789,'张三',35,'1980/1/1',4000,101),
(1674,'李四',32,'1983/4/1',3500,101),
(1776,'王五',24,'1990/7/1',2000,101),
(1568,'赵六',57,'1970/10/11',7500,102),
(1564,'荣七',64,'1963/10/11',8500,102),
(1879,'牛八',55,'1971/10/20',7300,103)，
   (1668, '钱九', 64, '1963/5/4', 8000, 102)，
   (1724, '武十', 22, '2023/5/8', 1500, 103)，
   (1770, '孙二', 65, '1986/8/12', 9500, 101)，
   (18400, '苟一', 65, '1986/8/12', 1500, 101);

二.连接查询操作

1.找出销售部门中年纪最大的员工的姓名

select e.name from emp e
join dept d
on e.dept2=d.dept1
order by e.age desc
limit 1;

2.求财务部门最低工资的员工姓名

select e.name from emp e
join dept d on e.dept2=d.dept1
where d.dept_name='财务'
order by e.incoming ASC
limit 1;

3.列出每个部门收入总和高于9000的部门名称

SELECT d.dept_name, SUM(e.incoming) AS total_income
FROM dept d
JOIN emp e ON d.dept1 = e.dept2
GROUP BY d.dept_name
HAVING total_income > 9000;

4.求工资在7500到8500元之间，年龄最大的人的姓名及部门

SELECT e.name, d.dept_name FROM emp e
JOIN dept d ON e.dept2 = d.dept1
WHERE e.incoming BETWEEN 7500 AND 8500
ORDER BY e.age DESC
LIMIT 1;

5.找出销售部门收入最低的员工入职时间

SELECT e.worktime_start
FROM emp e
JOIN dept d ON e.dept2 = d.dept1
WHERE d.dept_name = '销售'
ORDER BY e.incoming ASC
LIMIT 1;

6.财务部门收入超过2000元的员工姓名

SELECT e.name
FROM emp e
JOIN dept d ON e.dept2 = d.dept1
WHERE d.dept_name = '财务' AND e.incoming > 2000;

7.列出每个部门的平均收入及部门名称

SELECT d.dept_name, AVG(e.incoming) AS average_income
FROM dept d
JOIN emp e ON d.dept1 = e.dept2
GROUP BY d.dept_name;

8.IT技术部入职员工的员工号

select e.sid from emp e
join dept d on e.dept2=d.dept1
where d.dept_name='IT技术'；

9.财务部门的收入总和；

select sum(e.incoming)
from emp e
join dept d on e.dept2=d.dept1
where d.dept_name='财务';

10.找出哪个部门还没有员工入职；

SELECT d.dept_name FROM dept d LEFT JOIN emp e ON d.dept1 = e.dept2 WHERE e.sid IS NULL;

11.列出部门员工收入大于7000的部门编号，部门名称；

select d.dept1,d.dept_name from dept d
-> join emp e on d.dept1=e.dept2
-> where e.incoming>7000;

12.列出每一个部门的员工总收入及部门名称；

SELECT d.dept_name, SUM(e.incoming) AS total_income

FROM dept d

JOIN emp e ON d.dept1 = e.dept2

GROUP BY d.dept_name;

13.列出每一个部门中年纪最大的员工姓名，部门名称；

SELECT e.name, d.dept_name FROM emp e INNER JOIN dept d ON e.dept2 = d.dept1 WHERE (e.dept2, e.age) IN (SELECT dept2, MAX(age) FROM emp GROUP BY dept2);

14.求李四的收入及部门名称

SELECT e.incoming, d.dept_name FROM emp e INNER JOIN dept d ON e.dept2 = d.dept1 WHERE e.name = '李四';

15.列出每个部门中收入最高的员工姓名，部门名称，收入，并按照收入降序

SELECT e.name, d.dept_name, e.incoming FROM emp e INNER JOIN dept d ON e.dept2 = d.dept1 WHERE (e.dept2, e.incoming) IN (SELECT dept2, MAX(incoming) FROM emp GROUP BY dept2) ORDER BY e.incoming DESC;