Database multi-table connection query exercise
1. Create database dept, emp
2. Insert data
1. Find the name of the oldest employee in the sales department
mysql> select name,age from emp
inner join dept on dept.dept1=emp.dept2
where dept_name='销售'
order by age desc limit 1;
2. Find the name of the employee with the minimum salary in the financial department
mysql> select name from emp
where incoming=(select min(incoming) from emp);
3. List the names of the departments whose total income of each department is higher than 9000
mysql> select dept_name as 部门名称,sum(incoming) as 总工资 from emp
inner join dept on emp.dept2=dept.dept1
group by dept_name having 总工资>9000;
4. Ask for the salary between 7500 and 8500 yuan, the name and department of the oldest person
mysql> select name from emp
where incoming in (7500,8500)
order by age desc limit 1;
5. Find out when the lowest-paid employees in the sales department joined
mysql> select worktime_start from emp
inner join dept on emp.dept2=dept.dept1
where dept_name='销售'
order by incoming limit 1;
6. Names of employees in the financial department whose income exceeds 2,000 yuan
mysql> select name as 姓名 from emp
inner join dept on emp.dept2=dept.dept1
where dept_name='财务' and incoming>2000;
7. List the average income of each department and the name of the department
mysql> select dept_name as 部门名称, avg(incoming) as 平均工资 from emp
-> inner join dept on emp.dept2=dept.dept1
-> group by dept_name;
8. The employee number of the employee who joined the IT technology department
mysql> select sid from emp
-> inner join dept on emp.dept2=dept.dept1
-> where dept_name='IT技术';
9. The sum of the income of the financial department;
mysql> select sum(incoming) as 收入总和 from emp
-> inner join dept on emp.dept2=dept.dept1
-> where dept_name='财务';
10. First sort by the size of the department number, and then sort the employee information table from early to late according to the entry time
mysql> select * from emp order by dept2 desc,worktime_start;
11. Find out which department has no employees yet;
mysql> select dept_name from emp right join dept on emp.dept2=dept.dept1 where sid is null;
12. List the department numbers and department names of department employees whose income is greater than 7000;
mysql> select dept2 as 部门编号,dept_name as 部门名称 from emp
-> inner join dept on emp.dept2=dept.dept1
-> where incoming>7000;
13. List the total employee income and department name of each department;
mysql> select dept_name as 部门名称,sum(incoming) as 总收入 from emp -> inner join dept on emp.dept2=dept.dept1 -> group by dept_name;
14. List the name and department name of the oldest employee in each department;
mysql> select emp.name as 姓名, dept.dept_name as 部门名称 from dept
-> inner join (select max(age) as age,dept2 from emp group by dept2) as em on dept.dept1=em.dept2
-> inner join emp on dept.dept1=emp.dept2 and em.age=emp.age;
15. Find Li Si's income and department name
mysql> select incoming as 收入,dept_name from emp
-> inner join dept on emp.dept2=dept.dept1
-> where name='李四';
16. List the names of employees with the highest income in each department, department name, income, and in descending order of income
mysql> select name as 姓名,incoming as 收入,dept_name as 部门名称 from emp
-> inner join dept on emp.dept2=dept.dept1 -> inner join
-> (select max(incoming) as max_incom, dept2 from emp group by dept2)
-> as em
-> on emp.dept2=em.dept2 and emp.incoming=em.max_incom
-> order by incoming desc;
17. List the names of departments with more than 1 employee
mysql> select dept_name as 部门名称 from dept
-> inner join (select count(*) as 员工数,dept2 from emp group by dept2) -> as de on dept.dept1=de.dept2
-> where de.员工数>1;
19. Find the name of Zhang San's department
mysql> select dept_name as 部门名称 from emp
-> inner join dept on emp.dept2=dept.dept1
-> where name='张三';
