常见MySQL面试题解析一
首先申明一下要用的表
员工表
EMPLOYEE_ID：员工ID
FIRST_NAME：姓
LAST_NAME：名
EMAIL：email
phone_int：电话
HIRE_DATE：雇佣时间
JOB_ID：工作id
SALARY：月薪e
COMMISSION_PCT：奖金点
MANAGER_ID：该员工经理ID
DEPARTMENT_ID：该员工所属部门id
部门表
DEPARTMENT_ID：部门ID
DEPARTMENT_NAME：部门名称
MANAGER_ID：部门管理员ID
LOCATION_ID：部门地址ID
地址表：
LOCATION_ID：地址ID
STREET_ADDRESS：地址信息
POSTAL_CODE：邮政编码
CITY：城市
STATE_PROVINCE：省份
COIUNTRY_ID：所属国家ID
国家表：
COUNTRY_ID：国家id
COUNTRY_NAME：国家名字
REGION_ID：所属地域ID
地域表：
REGION_ID：地域ID
REGION_NAME：地域名称
员工工作历史表：
EMPLOYEE_ID：员工ID
START_DATE：开始工作时间
END_DATE：工作结束时间
JOB_ID：工作ID
DEPARTMENT_ID：所属部门ID
工资等级表：
GRADE_LEVEL：等级；
LOWEST_SAL：该等级下限；
HIGHEST_SAL：该等级上限；
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一、查询员工的全名，email和电话
select
CONCAT(first_name,last_name) full_name,
employee_id,email,phone_int
from employees
string拼接函数concat的使用
二、查询所有员工的全名，月薪和年薪（月薪*12）
select
employee_id,CONCAT(first_name,last_name) full_name,
salary, salary * 12 year_income
from employees
考点：涉及到列的运算
三、查询所有员工的全名，月薪和年终奖(年薪*commission_pct)
select
CONCAT(first_name,last_name) full_name,salary monthIncome,
salary*12*ifnull(commission_pct,0) yearReward
from employees
这里有坑，因为commission_pct完全可能为空
此处考察对于null的处理，ifnull(exp,defaultValue),若exp为null则采用默认值
四、哪些部门有员工,员工的职位是什么
SELECT
count(e.email),e.job_id
from employees e right join departments d
on e.department_id = d.department_id
五、查询1999年之后(含1999年)入职的员工信息
select * from employees where `YEAR`(HIRE_DATE) = 1999;
考察MySQL内置函数的用法
六、查询公司的老板信息
SELECT *
from employees
where manager_id is null
此处考察对于列值为null的判断,老板由谁管？
七、查询所有员工信息，按照部门升序和年薪降序排序；
select
CONCAT(first_name,last_name) full_name,
salary*12 yearIncome , DEPARTMENT_ID
from employees
order by DEPARTMENT_ID asc , yearIncome desc;
八、计算员工姓名和全薪；
select CONCAT(first_name,last_name) full_name,
(salary * 12 + salary * 12 * IFNULL(commission_pct,0)) fullSalary
from employees
这里同样是个大坑，具体考察对于null的处理,null和谁发生运算结果都是null
九、查询员工所属的部门id，如果没有部门，打印“未分配部门”
select e.employee_id, ifnull(e.department_id,"未分配部门")
from employees
对于null值的优雅处理
十、查询出每一个部门的平均工资
select d.department_id, avg(e.salary)
from employees e join departments d on e.department_id = d.department_id
group by e.department_id
分组的考察
十一、查询平均工资高于8000的部门和其平均工资
select d.department_id, d.department.name
from employees e join departments d on e.department_id = d.department_id
group by e.department_id having avg(e.salary) > 8000
group by + having 的考察，手写的时候容易忘记，having必须在有group的前提下才能使用
十二、查询换过工作员工换工作的次数
select e.employee_id, CONCAT(first_name,last_name) full_name,count(e.employee_id)
from employees e join job_history jh on jh.employee_id = e.employee_id
group by e.employee_id
十三、查询在1995，1996，1997，1998年各进公司多少人
select Year(hire_date),count(*)
from employees
where Year(hire_date) >= '1995' and Year(hire_date) <= '1998'
group by Year(hire_date)
对于in的考察
十四、输出员工名称，员工id，员工所属部门
select e.employee_id, CONCAT(first_name,last_name) full_name, d.department_name
from employees e join departments d on e.department_id = d.department_id
十五、输出员工信息，包括employee_id, first_name, department_id,
department_name location_id city
select employee_id,first_name,last_name,
d.DEPARTMENT_NAME,d.department_id,city
from employees e INNER JOIN departments d
on e.DEPARTMENT_ID = d.DEPARTMENT_ID
INNER JOIN locations lo on lo.location_id = d.location_id
十六、在 EMPLOYEES 表中薪水的最低级别和最高级别分别是什么？
select jg.grade_level from job_grades jg join
(
select salary from employees where
salary = (select MIN(salary) from employees ) or salary = (select MAX(salary) from employees)
)t
on t.salary BETWEEN jg.lowest_sal and jg.highest_sal
十七、查询所有有奖金的员工的姓名，部门，地址，城市
select
e.employee_id,CONCAT(first_name,last_name) full_name,
d.department_name,
lo.street_address,lo.city,
c.country_name,
r.region_name
from
employees e join departments d on e.department_id = d.department_id
join locations lo on lo.location_id = d.location_id
join countries c on c.country_id = lo.country_id
join regions r on r.region_id = c.region_id
where e.commission_pct is not null
十七、查询last_name为’Ki’ 的员工的经理信息
select m.employee_id,CONCAT(m.first_name,m.last_name) full_name
from employees e join employees m on e.manager_id = m.employee_id
where e.last_name like concat('Ki','%')
十八、查询公司工资最低的员工信息
select employee_id,CONCAT(first_name,last_name) full_name, min(salary)
from employees
十九、查询1999年进入公司的员工的最高工资的员工
select e.employee_id,CONCAT(first_name,last_name) full_name, max(e.salary)
from employees e where YEAR(e.hire_date) = '1999'
二十、查询曾经做过ST_CLERK的员工信息
select
e.employee_id, CONCAT(first_name,last_name) full_name
from employees e join job_history jh on e.employee_id = jh.employee_id
where jh.job_id = 'ST_CLERK' and e.job_id != 'ST_CLERK'
胜负统计问题：
select
date,
(select count(*) from game where result = '胜' and tab1.date = date),
(select count(*) from game where result = '负' and tab1.date = date)
from game tab1
group by date