SQL（01-20）

01. 查找最晚入职员工的所有信息

CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));

select * from employees 
where hire_date = (select max(hire_date) from employees);

-- 最晚的数据可能不止一条，数据集不对可能会错误
SELECT * FROM employees ORDER BY hire_date DESC LIMIT 0,1;

LIMIT m,n : 表示从第m+1条开始，取n条数据；

LIMIT n ： 表示从第0条开始，取n条数据，是limit(0,n)的缩写。

02. 查找入职员工时间排名倒数第三的员工所有信息

select * from employees order by hire_date DESC limit 2,1

-- 考虑了入职时间重复和人数重复的情况
select *
from employees
where hire_date=(
select distinct hire_date from employees order by hire_date desc limit 2,1);

03. 查找各个部门当前(to_date=‘9999-01-01’)领导当前薪水详情以及其对应部门编号dept_no

CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

select s.emp_no,s.salary,s.from_date,s.to_date,d.dept_no 
from salaries s left join dept_manager d
on d.emp_no = s.emp_no 
where d.to_date = '9999-01-01' 
and s.to_date = '9999-01-01';

04. 查找所有已经分配部门的员工的last_name和first_name

CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));

• left join(左联接)

返回包括左表中的所有记录和右表中联结字段相等的记录，左表的记录将会全部表示出来,而右表条件不满足的地方均为NULL.

• right join(右联接)

返回包括右表中的所有记录和左表中联结字段相等的记录，左表不符合的都为NULL

• inner join(等值连接) 只返回两个表中联结字段相等的行

select e.last_name,e.first_name,d.dept_no
from employees e inner join dept_emp d
on e.emp_no = d.emp_no

使用左连接查询时，employees中没有分配部门的员工（没有被记录在dept_emp表）dept_no字段被自动取NULL然后被输出，所以应当剔除（复合条件连接查询）。

select last_name,first_name,dept_no 
from employees left join dept_emp
on employees.emp_no = dept_emp.emp_no
where dept_emp.dept_no<>'';

05. 查找所有员工的last_name和first_name以及对应部门编号dept_no，也包括展示没有分配具体部门的员工

select last_name,first_name,dept_no 
from employees left join dept_emp
on employees.emp_no = dept_emp.emp_no

06. 查找所有员工入职时候的薪水情况，给出emp_no以及salary， 并按照emp_no进行逆序

CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

SELECT e.emp_no, s.salary
FROM employees e inner join salaries s
on e.emp_no = s.emp_no
where e.hire_date = s.from_date
order by e.emp_no DESC 

07. 查找薪水涨幅超过15次的员工号emp_no以及其对应的涨幅次数t

CREATE TABLE salaries (
emp_no int(11) NOT NULL,
salary int(11) NOT NULL,
from_date date NOT NULL,
to_date date NOT NULL,
PRIMARY KEY (emp_no,from_date));

SELECT emp_no,count(*) as t
FROM salaries
GROUP BY emp_no
HAVING t>15

GROUP BY…HAVING
由于COUNT()函数不可用于WHERE语句中，故使用HAVING语句来限定t>15的条件

08. 找出所有员工当前(to_date=‘9999-01-01’)具体的薪水salary情况，对于相同的薪水只显示一次,并按照逆序显示

CREATE TABLE salaries (
emp_no int(11) NOT NULL,
salary int(11) NOT NULL,
from_date date NOT NULL,
to_date date NOT NULL,
PRIMARY KEY (emp_no,from_date));

select distinct(salary) as salary
from salaries
where to_date='9999-01-01'
order by salary desc

09. 获取所有部门当前manager的当前薪水情况，给出dept_no, emp_no以及salary，当前表示to_date=‘9999-01-01’

CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

select d.dept_no,d.emp_no,s.salary
from dept_manager d left join salaries s
on d.emp_no = s.emp_no
where d.to_date='9999-01-01'
and s.to_date='9999-01-01'

10. 获取所有非manager的员工emp_no

CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));

方法一：使用NOT IN选出在employees但不在dept_manager中的emp_no记录

扫描二维码关注公众号，回复： 5721304 查看本文章

SELECT emp_no FROM employees
WHERE emp_no NOT IN (SELECT emp_no FROM dept_manager)

方法二：先使用LEFT JOIN连接两张表，再从此表中选出dept_no值为NULL对应的emp_no记录

SELECT emp_no FROM (SELECT * FROM employees LEFT JOIN dept_manager
ON employees.emp_no = dept_manager.emp_no)
WHERE dept_no IS NULL

方法三：方法二的简版，使用单层SELECT语句即可

SELECT employees.emp_no FROM employees LEFT JOIN dept_manager
ON employees.emp_no = dept_manager.emp_no
WHERE dept_no IS NULL

11. 获取所有员工当前的manager，如果当前的manager是自己的话结果不显示，当前表示to_date=‘9999-01-01’。

结果第一列给出当前员工的emp_no,第二列给出其manager对应的manager_no。

CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `dept_manager` (
`dept_no` char(4) NOT NULL,
`emp_no` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));

select de.emp_no as emp_no,
        dm.emp_no as manager_no
from dept_emp de inner join dept_manager dm
on de.dept_no = dm.dept_no
where de.to_date='9999-01-01' and dm.to_date='9999-01-01'
and de.emp_no<>dm.emp_no 

12. 获取所有部门中当前员工薪水最高的相关信息，给出dept_no, emp_no以及其对应的salary

CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

select d.dept_no,s.emp_no,max(s.salary) as salary
from dept_emp d inner join salaries s
on d.emp_no = s.emp_no
WHERE d.to_date = '9999-01-01' AND s.to_date = '9999-01-01'
group by d.dept_no

13. 从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。

CREATE TABLE IF NOT EXISTS "titles" (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);

select  title, count(*) as t
from titles
group by title
having t>=2

14. 从titles表获取按照title进行分组，每组个数大于等于2，给出title以及对应的数目t。

注意对于重复的emp_no进行忽略。

CREATE TABLE IF NOT EXISTS "titles" (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);

select  title, count(distinct(emp_no)) as t
from titles
group by title
having t>=2

15. 查找employees表所有emp_no为奇数，且last_name不为Mary的员工信息，并按照hire_date逆序排列

CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));

select *
from employees
where emp_no%2=1 and last_name<>'Mary'
order by hire_date DESC

16. 统计出当前各个title类型对应的员工当前薪水对应的平均工资。结果给出title以及平均工资avg。

CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
CREATE TABLE IF NOT EXISTS "titles" (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);

select t.title,avg(s.salary) as avg
from titles t inner join salaries s 
on t.emp_no = s.emp_no
where s.to_date = '9999-01-01' and t.to_date = '9999-01-01'
group by t.title

17. 获取当前（to_date=‘9999-01-01’）薪水第二多的员工的emp_no以及其对应的薪水salary

CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

select emp_no,salary
from salaries
where to_date='9999-01-01' and
salary in(
    select distinct(salary)
    from salaries
    where to_date='9999-01-01'
    order by salary desc
    limit 1,1
)

18. 查找当前薪水(to_date=‘9999-01-01’)排名第二多的员工编号emp_no、薪水salary、last_name以及first_name，不准使用order by

CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

select e.emp_no,max(s.salary),e.last_name,e.first_name
from employees e inner join salaries s
on e.emp_no = s.emp_no
where s.to_date='9999-01-01' --只需要限制当前薪水
and s.salary<(
select max(salary)
from salaries
where to_date='9999-01-01'
)

19. 查找所有员工的last_name和first_name以及对应的dept_name，也包括暂时没有分配部门的员工

CREATE TABLE `departments` (
`dept_no` char(4) NOT NULL,
`dept_name` varchar(40) NOT NULL,
PRIMARY KEY (`dept_no`));
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));

select e.last_name,e.first_name,d.dept_name
from (employees e left join dept_emp de
on e.emp_no = de.emp_no) 
left join departments d
on de.dept_no = d.dept_no

20. 查找员工编号emp_no为10001其自入职以来的薪水salary涨幅值growth

CREATE TABLE `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));

select max(salary)-min(salary) as growth
from salaries
where emp_no = 10001

SELECT ( 
(SELECT salary FROM salaries WHERE emp_no = 10001 ORDER BY to_date DESC LIMIT 1) -
(SELECT salary FROM salaries WHERE emp_no = 10001 ORDER BY to_date ASC LIMIT 1)
) AS growth