MySQL的查询练习

– 基础查询练习
显示出表employees部门编号在80-100之间 的姓名、职位

select last_name,job_id from employees
where department_id >= 80 and department_id <=100

显示出表employees的manager_id 是 100,101,110 的员工姓名、职位

select last_name,job_id from employees
WHERE manager_id=100 OR manager_id=101 OR manager_id = 110;
select last_name,job_id from employees
where manager_id in(100,101,110)

#案例1：查询姓名中包含字符 e的员工信息

select * from employees
where last_name like '%e%'

#案例2：查询姓名中第二个字符为e，第四个字符为a的员工信息

select * from employees
where last_name like '_e_a%'

#案例1：查询没有奖金的员工

select * from employees
where commission_pct is null 
//有奖金
select *from employees
where commission_pct is not null 

查询员工号为176的员工的姓名和部门号和年薪

select last_name,department_id,salary*12*(1+IFNULL(commission_pct,0))
from employees
where employee_id=176

选择工资不在5000到12000的员工的姓名和工资

SELECT last_name,salary from employees
where  not(salary BETWEEN 5000 and 12000)

选择在20或50号部门工作的员工姓名和部门号

SELECT last_name,department_id from employees
where department_id in(20,50)

按单个字段进行排序

SELECT * from employees
order by salary asc 
SELECT *FROM employees
ORDER BY salary DESC

按多个字段进行排序

select * from employees
ORDER BY salary desc,employee_id ASC

按表达式排序
#案例：按年薪降序

SELECT salary*12*(1+IFNULL(commission_pct,0)) 年薪 from employees
order by 年薪 DESC

按函数排序

#案例：按姓名中的字节长度大小降序
select last_name,LENGTH(last_name) from employees
order by LENGTH(last_name) DESC
#1.查询员工的姓名和部门号和年薪，按年薪降序 按姓名升序
SELECT last_name,department_id,salary*12*(1+IFNULL(commission_pct,0)) 年薪 
from employees
ORDER BY 年薪 DESC,last_name ASC
#2.选择工资不在8000到17000的员工的姓名和工资，按工资降序
select last_name,salary from employees
where salary not BETWEEN 8000 and 17000
ORDER BY salary DESC
#3.查询邮箱中包含e的员工信息，并先按邮箱的字节数降序，再按部门号升序
SELECT * FROM employees
where email like'%e%'
ORDER BY LENGTH(email) DESC,department_id asc

#01案例：查询每个部门的员工个数和部门名

SELECT COUNT(*) 个数,department_name
from employees e,departments d
where e.employee_id=d.department_id
GROUP BY e.department_id
having 个数>5

#案例2：查询名字中第三个字符为a，第五个字符为e的员工的工资以及对应的工资级别

select salary,grade_level from employees e,job_grades g
where e.salary BETWEEN lowest_sal and highest_sal
and e.last_name like '__a_e%'

#案例2：查询有奖金的员工名、部门名

select last_name,department_name 
from departments d
join employees e on e.department_id=d.department_id
where commission_pct is not NULL

#案例3：查询城市名、员工名和部门名

select city,last_name,department_name from employees e
 inner join departments d on e.department_id=d.department_id
 inner join locations l on l.location_id=d.location_id

查询各部门中工资比本部门平均工资高的员工的员工号, 姓名和工资

#查询各个部门的平均工资
select AVG(salary),department_id from employees
where department_id is not null 
group by department_id 
#查询员工 工号 姓名 工资
select salary,last_name,employee_id,e.department_id 
FROM employees e,
(
SELECT department_id,AVG(salary) 部门平均工资
	FROM employees
	GROUP BY department_id
)f
where e.department_id = f.department_id and e.salary > f.部门平均工资

查询工资比公司平均工资高的员工的员工号，姓名和工资。

select avg(salary) from employees

select employee_id,last_name,salary from employees
where salary>(
select avg(salary) from employees
)

查询姓名中包含字母u的员工在相同部门的员工的员工号和姓名

select department_id from employees
where last_name like '%u%'

select employee_id,last_name from employees
where department_id in(
select department_id from employees
where last_name like '%u%'
)

查询在部门的location_id为1700的部门工作的员工的员工号

select department_id from departments
where location_id = 1700

select employee_id from employees
where department_id in(
select department_id from departments
where location_id = 1700
)

查询管理者是King的员工姓名和工资

select employee_id from employees
where last_name ='King'

select last_name,salary from employees
where manager_id in(
select employee_id from employees
where last_name ='King'
)

查询平均工资最低的部门信息

select avg(salary) f,department_id from employees
group by department_id

select min(f) from (
select avg(salary) f,department_id from employees
group by department_id
)a

select avg(salary),department_id from  employees
group by department_id
having AVG(salary)=(
select min(f) from (
select avg(salary) f ,department_id from employees
group by department_id
)a
)
select * from departments 
where department_id=(
select department_id from  employees
where department_id is not null
group by department_id
having AVG(salary)=(
select min(f) from (
select avg(salary) f ,department_id from employees
group by department_id
)a
)
)