SQL专题1

一、牛客SQL题1

1、查找最晚入职员工的所有信息

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)

2、查找入职员工时间排名倒数第三的员工所有信息

题目描述
查找入职员工时间排名倒数第三的员工所有信息
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
order by hire_date desc limit 2,1

• limit 2,1 限制从第二开始，选第三个
• 含义是跳过2条取出1条数据，limit后面是从第2条开始读，读取1条信息，即读取第3条数据

select *
from employees
order by hire_date desc limit 1 offset 2

• 含义是从第1条（不包括）数据开始取出2条数据，limit后面跟的是2条数据，offset后面是从第1条开始读取，即读取第2,3条
• 跳出2，然后取1 limit

3、查找当前薪水详情以及部门编号dept_no

题目描述
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 salaries.*,dept_no
from salaries,dept_manager
where salaries.emp_no = dept_manager.emp_no and salaries.to_date='9999-01-01' and dept_manager.to_date='9999-01-01'

• 注意当前薪水，领导编号

4、查找所有已经分配部门的员工的last_name和first_name

题目描述
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));

select e.last_name,e.first_name,de.dept_no
from employees e,dept_emp de
where e.emp_no = de.emp_no

• 本题做内链接，就可以知道领导的薪水信息了

5、查找所有员工的last_name和first_name以及对应部门编号dept_no

题目描述
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
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,de.dept_no
from employees e left join dept_emp de
on e.emp_no = de.emp_no

• 本题主要考察左链接

6、查找所有员工入职时候的薪水情况

题目描述
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 salaries.emp_no ,salary
FROM salaries,employees
WHERE salaries.emp_no = employees.emp_no and employees.hire_date = salaries.from_date
ORDER BY salaries.emp_no desc

• 本题主要是员工的employees.hire_date = salaries.from_date

7、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

题目描述
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 函数的使用

8、查找所有员工当前薪水salary的情况

题目描述
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
from salaries
where to_date = '9999-01-01' 
order by salary Desc

• 注意distinct去重

9、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’

题目描述
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 dm.dept_no,dm.emp_no,s.salary
from dept_manager dm,salaries s
where dm.emp_no = s.emp_no and dm.to_date = "9999-01-01" and s.to_date = "9999-01-01"

• 主要是考虑日期的时间

10、获取所有非manager的员工emp_no

题目描述
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));

select emp_no
from employees
where emp_no not in (
select emp_no from dept_manager)

• 使用not in来筛选

11、获取所有员工当前的manager

题目描述
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,dm.emp_no as manager_no
from dept_emp de,dept_manager dm
where de.dept_no = dm.dept_no
and de.to_date = "9999-01-01" and dm.to_date = "9999-01-01"
and de.emp_no != dm.emp_no

• 本文主要考虑$如果当前的manager是自己的话结果不显示$
• 所以有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

题目描述
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 de.dept_no,de.emp_no,salary
from dept_emp de,salaries s
where de.emp_no = s.emp_no and de.to_date  = '9999-01-01' and s.to_date = '9999-01-01'
group by de.dept_no
having salary = max(salary)

select de.dept_no,de.emp_no,max(salary)
from dept_emp de,salaries s
where de.emp_no = s.emp_no and de.to_date  = '9999-01-01' and s.to_date = '9999-01-01'
group by de.dept_no

• sql里groupby 函数，好像只能显示聚和数，如果是单列数可能会报错？但是本题却运行成功？

select  dept_no,emp_no,salary
from 
(select d.dept_no,s.emp_no,s.salary,row_number() over(partition by d.dept_no order by salary desc) as rank
from dept_emp d,salaries s
where d.emp_no = s.emp_no
and d.to_date = '9999-01-01' and s.to_date = '9999-01-01') as a
where rank = 1

• 使用row_number来完成。

select de.dept_no, de.emp_no, s.salary
from dept_emp de inner join salaries s
on de.emp_no = s.emp_no
and de.to_date = '9999-01-01'
and s.to_date = '9999-01-01'
where s.salary in (select max(s2.salary)
from dept_emp de2 inner join salaries s2
on de2.emp_no = s2.emp_no
and de2.to_date = '9999-01-01'
and s2.to_date = '9999-01-01'
where de2.dept_no = de.dept_no
group by de2.dept_no)
order by de.dept_no

• 这种写法好像mysql不支持

链接：https://www.nowcoder.com/questionTerminal/4a052e3e1df5435880d4353eb18a91c6?f=discussion
来源：牛客网

SELECT e.dept_no, e.emp_no, s.salary
FROM dept_emp AS e INNER JOIN salaries AS s
ON e.emp_no = s.emp_no
WHERE e.to_date = '9999-01-01' AND s.to_date = '9999-01-01'
GROUP BY e.dept_no
HAVING s.salary = MAX(s.salary);

13、从titles表获取按照title进行分组

题目描述
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 count(*) >=2 

14、从titles表获取按照title进行分组，注意对于重复的emp_no进行忽略。

题目描述
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title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

• 本题主要是对 count(distinct emp_no)进行筛选

15、查找employees表

题目描述
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 last_name != "Mary" and emp_no%2 = 1
order by hire_date desc

• 本题主要!=‘MARY’ and emp_no % 2 =1

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

题目描述
统计出当前各个title类型对应的员工当前（to_date=‘9999-01-01’）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 title,avg(salary) as avg
from salaries,titles
where salaries.emp_no = titles.emp_no
and salaries.to_date = '9999-01-01'
and titles.to_date = '9999-01-01'
group by title

• 主要avg

17、获取当前薪水第二多的员工的emp_no以及其对应的薪水salary

题目描述
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"
order by salary desc
limit 1,1

• 本题主要order by 然后 limit
• 也可以使用这个

select emp_no,salary
from salaries
where to_date = "9999-01-01"
order by salary desc
limit 1 offset 1

18、获取当前薪水第二多的员工的emp_no以及其对应的薪水salary，不准使用order by

题目描述
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(salary) as salary,last_name,first_name
from employees e,salaries s
where e.emp_no = s.emp_no
and s.to_date = "9999-01-01"
and salary <(select max(salary) from salaries where to_date = "9999-01-01") 

• 排名第二，说明现在的salary都小于 max(salary)
• 然后在该表中查询最大值（也就是原表中第二大的值）

19、查找所有员工的last_name和first_name以及对应的dept_name

题目描述
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,p.dept_name
from employees as e left join dept_emp d
on e.emp_no = d.emp_no left join departments p
on d.dept_no = p.dept_no

• 三表，主要是left join

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

题目描述
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
(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 limit 1)
as growth

-- 当前的值 减去 入职的值

• 当前时间的薪水 减去 入职时间的薪水（对入职时间排序，最小的就是 入职时间）

21、查找所有员工自入职以来的薪水涨幅情况

题目描述
查找所有员工自入职以来的薪水涨幅情况，给出员工编号emp_no以及其对应的薪水涨幅growth，并按照growth进行升序
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,(s1.salary-s2.salary) as growth
from employees e,salaries s1,salaries s2
where e.emp_no = s1.emp_no and s1.to_date = '9999-01-01' and  e.emp_no = s2.emp_no and s2.from_date = e.hire_date
order by growth 

• s1表是现在的薪水，s2表是入职的薪水，两者的时间不同。

22、统计各个部门对应员工涨幅的次数总和

题目描述
统计各个部门对应员工涨幅的次数总和，给出部门编码dept_no、部门名称dept_name以及次数sum
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 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 de.dept_no,dt.dept_name,count(s.salary) as sum
from dept_emp de,departments dt,salaries s
where de.emp_no = s.emp_no and dt.dept_no = de.dept_no
group by de.dept_no

• 薪水涨幅，和上一题思路类似

23、统计各个部门对应员工涨幅的次数总和

题目描述
统计各个部门对应员工涨幅的次数总和，给出部门编码dept_no、部门名称dept_name以及次数sum
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 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 de.dept_no,dt.dept_name,count(s.salary) as sum
from dept_emp de,departments dt,salaries s
where de.emp_no = s.emp_no and dt.dept_no = de.dept_no
group by de.dept_no,dt.dept_name

24、对所有员工的薪水按照salary进行按照1-N的排名

题目描述
对所有员工的当前(to_date=‘9999-01-01’)薪水按照salary进行按照1-N的排名，相同salary并列且按照emp_no升序排列
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 s1.emp_no, s1.salary, COUNT(distinct s2.salary) AS rank
FROM salaries AS s1, salaries AS s2
WHERE s1.to_date = '9999-01-01' AND s2.to_date = '9999-01-01' AND s1.salary <= s2.salary
GROUP BY s1.emp_no
ORDER BY s1.salary DESC, s1.emp_no

select emp_no,salary, dense_rank() over(order by salary desc) as rank
from salaries
where to_date = '9999-01-01'

• 使用dense_rank() 排名

select s1.emp_no,s1.salary,count(distinct s2.salary) as rank
from salaries s1,salaries s2
where s1.to_date = '9999-01-01' and s2.to_date = '9999-01-01'
and s1.salary <= s2.salary
group by s1.emp_no
order by s1.salary desc

• s1.salary降序排列，然后s2.salary 都要大于等于s1.salary，就有次数了

• distinct是由于 rank里有重复的排名

25、获取所有非manager员工当前的薪水情况

题目描述
获取所有非manager员工当前的薪水情况，给出dept_no、emp_no以及salary ，当前表示to_date=‘9999-01-01’
CREATE TABLE dept_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));
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 de.dept_no,e.emp_no,s.salary
from employees e,salaries s,dept_emp de
where e.emp_no = s.emp_no and e.emp_no = de.emp_no and s.to_date = "9999-01-01" and  
de.to_date = '9999-01-01' and
e.emp_no not in 
(select emp_no from dept_manager)

• 本题主要not in
• 其他做链接

26、获取员工其当前的薪水比其manager当前薪水还高的相关信息

题目描述
获取员工其当前的薪水比其manager当前薪水还高的相关信息，当前表示to_date=‘9999-01-01’,
结果第一列给出员工的emp_no，
第二列给出其manager的manager_no，
第三列给出该员工当前的薪水emp_salary,
第四列给该员工对应的manager当前的薪水manager_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 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 a.emp_no emp_no, b.emp_no manager_no,a.salary emp_salary,b.salary manager_salary
from 
(select  * from dept_emp de,salaries s 
where de.emp_no = s.emp_no and de.to_date = "9999-01-01" and s.to_date = "9999-01-01") as a
,
(select * from dept_manager dm, salaries s
where dm.emp_no = s.emp_no and dm.to_date = "9999-01-01" and s.to_date = "9999-01-01") as b
where a.salary > b.salary and a.dept_no = b.dept_no 

select a.emp_no,b.emp_no as manager_no,a.salary as emp_salary,b.salary as manager_salary
from 
(select de.*,s.salary
from dept_emp de,salaries s
where de.emp_no = s.emp_no and s.to_date ='9999-01-01') as a,
(select dm.*,s.salary
from dept_manager dm,salaries s
where dm.emp_no = s.emp_no and s.to_date = '9999-01-01') as b
where a.dept_no = b.dept_no and a.salary > b.salary

• 建立员工薪资表和经理薪资表
• 然后按部门编号做链接，然后比较薪资，重命名

27、汇总各个部门当前员工的title类型的分配数目

题目描述
汇总各个部门当前员工的title类型的分配数目，结果给出部门编号dept_no、dept_name、其当前员工所有的title以及该类型title对应的数目count
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 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 d.dept_no,d.dept_name,t.title,count(*) as count
from departments d,dept_emp de,titles t
where d.dept_no = de.dept_no and t.emp_no = de.emp_no and
t.to_date = '9999-01-01' and de.to_date = '9999-01-01'
group by d.dept_no,d.dept_name,t.title

28、给出每个员工每年薪水涨幅超过5000的员工编号emp_no

题目描述
给出每个员工每年薪水涨幅超过5000的员工编号emp_no、薪水变更开始日期from_date以及薪水涨幅值salary_growth，并按照salary_growth逆序排列。
提示：在sqlite中获取datetime时间对应的年份函数为strftime(’%Y’, to_date)

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 s1.emp_no, s2.from_date, s2.salary - s1.salary AS salary_growth
FROM salaries AS s1, salaries AS s2
WHERE s1.emp_no = s2.emp_no 
AND strftime('%Y', s2.to_date) - strftime('%Y', s1.to_date) = 1 
AND salary_growth > 5000
ORDER BY salary_growth DESC;

• 建立两表salaries,做比较

29、查找描述信息中包括robot的电影对应的分类名称以及电影数目，而且还需要该分类对应电影数量>=5部

题目描述
film表
字段 说明
film_id 电影id
title 电影名称
description 电影描述信息

CREATE TABLE IF NOT EXISTS film (
film_id smallint(5) NOT NULL DEFAULT ‘0’,
title varchar(255) NOT NULL,
description text,
PRIMARY KEY (film_id));
category表
字段 说明
category_id 电影分类id
name 电影分类名称
last_update 电影分类最后更新时间

CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, last_update timestamp,
PRIMARY KEY ( category_id ));
film_category表
字段 说明
film_id 电影id
category_id 电影分类id
last_update 电影id和分类id对应关系的最后更新时间

CREATE TABLE film_category (
film_id smallint(5) NOT NULL,
category_id tinyint(3) NOT NULL, last_update timestamp);

查找描述信息中包括robot的电影对应的分类名称以及电影数目，而且还需要该分类对应电影数量>=5部

select c.name,count(fc.film_id)
from (select * from film where description like "%robot%") as f,
(select category_id,count(film_id) from film_category group by category_id having count(film_id) >=5) as cc,
category c,film_category fc
where f.film_id = fc.film_id and fc.category_id = c.category_id and cc.category_id = fc.category_id

• 首先筛选robot，然后筛选出 count(film_id) >= 5的表，然后再做链接

https://www.nowcoder.com/practice/3a303a39cc40489b99a7e1867e6507c5?tpId=82&tqId=29780&rp=0&ru=/ta/sql&qru=/ta/sql/question-ranking

• 需要注意

30、使用join查询方式找出没有分类的电影id以及名称

题目描述
film表
字段 说明
film_id 电影id
title 电影名称
description 电影描述信息

CREATE TABLE IF NOT EXISTS film (
film_id smallint(5) NOT NULL DEFAULT ‘0’,
title varchar(255) NOT NULL,
description text,
PRIMARY KEY (film_id));
category表
字段 说明
category_id 电影分类id
name 电影分类名称
last_update 电影分类最后更新时间

CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, last_update timestamp,
PRIMARY KEY ( category_id ));
film_category表
字段 说明
film_id 电影id
category_id 电影分类id
last_update 电影id和分类id对应关系的最后更新时间

CREATE TABLE film_category (
film_id smallint(5) NOT NULL,
category_id tinyint(3) NOT NULL, last_update timestamp);

使用join查询方式找出没有分类的电影id以及名称

select f.film_id,f.title
from film f 
left join film_category fc
on f.film_id = fc.film_id
left join category  c
on fc.category_id = c.category_id
where c.category_id is null

select film.film_id,film.title
from film left join film_category
on film.film_id = film_category.film_id
where category_id is null

• 主要是left join 两表链接 就可以了

31、使用子查询的方式找出属于Action分类的所有电影对应的title,description

题目描述
film表
字段 说明
film_id 电影id
title 电影名称
description 电影描述信息

CREATE TABLE IF NOT EXISTS film (
film_id smallint(5) NOT NULL DEFAULT ‘0’,
title varchar(255) NOT NULL,
description text,
PRIMARY KEY (film_id));
category表
字段 说明
category_id 电影分类id
name 电影分类名称
last_update 电影分类最后更新时间

CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, last_update timestamp,
PRIMARY KEY ( category_id ));
film_category表
字段 说明
film_id 电影id
category_id 电影分类id
last_update 电影id和分类id对应关系的最后更新时间

CREATE TABLE film_category (
film_id smallint(5) NOT NULL,
category_id tinyint(3) NOT NULL, last_update timestamp);

使用子查询的方式找出属于Action分类的所有电影对应的title,description

select f.title,f.description
from category c,film f,film_category fc
where c.name = 'Action'
and f.film_id = fc.film_id and fc.category_id = c.category_id

select title,description
from film 
where film_id
in (select film_id from film_category where category_id in 
   (select category_id from category where name = 'Action'))

select title,description
from film 
where film_id
in (select film_id from film_category  fc, category c 
    where name = 'Action' and fc.category_id = c.category_id)

32、获取select * from employees对应的执行计划

题目描述
获取select * from employees对应的执行计划

explain select * from employees

33、将employees表的所有员工的last_name和first_name拼接起来作为Name

题目描述
将employees表的所有员工的last_name和first_name拼接起来作为Name，中间以一个空格区分
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 last_name || ' ' || first_name as Name
from employees

34、创建一个actor表，包含如下列信息

题目描述
创建一个actor表，包含如下列信息
列表 类型 是否为NULL 含义
actor_id smallint(5) not null 主键id
first_name varchar(45) not null 名字
last_name varchar(45) not null 姓氏
last_update timestamp not null 最后更新时间，默认是系统的当前时间

create table if not exists actor(
actor_id smallint(5) not null,
first_name varchar(45) not null,
last_name varchar(45) not null,
last_update timestamp not null default (datetime('now','localtime')),
primary key(actor_id))

• 当前时间datetime(‘now’,‘localtime’)

35、批量插入数据

题目描述
对于表actor批量插入如下数据
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime(‘now’,‘localtime’)))
actor_id first_name last_name last_update
1 PENELOPE GUINESS 2006-02-15 12:34:33
2 NICK WAHLBERG 2006-02-15 12:34:33

insert into actor values
(1,'PENELOPE','GUINESS','2006-02-15 12:34:33'),
(2,'NICK','WAHLBERG','2006-02-15 12:34:33')

36、批量插入数据，不使用replace操作

题目描述
对于表actor批量插入如下数据,如果数据已经存在，请忽略，不使用replace操作
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime(‘now’,‘localtime’)))
actor_id first_name last_name last_update
‘3’ ‘ED’ ‘CHASE’ ‘2006-02-15 12:34:33’

INSERT OR IGNORE INTO actor VALUES (3, 'ED', 'CHASE', '2006-02-15 12:34:33')

• 主要是or ignore的使用

37、创建一个actor_name表

题目描述
对于如下表actor，其对应的数据为:
actor_id first_name last_name last_update
1 PENELOPE GUINESS 2006-02-15 12:34:33
2 NICK WAHLBERG 2006-02-15 12:34:33

创建一个actor_name表，将actor表中的所有first_name以及last_name导入改表。 actor_name表结构如下：
列表 类型 是否为NULL 含义
first_name varchar(45) not null 名字
last_name varchar(45) not null 姓氏

create table if not exists actor_name (
    first_name varchar(45) not null,
    last_name varchar(45) not null
);
insert into actor_name select first_name,last_name from actor;

create table actor_name as
select first_name,last_name from actor;

38、对first_name创建唯一索引uniq_idx_firstname

题目描述
针对如下表actor结构创建索引：
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime(‘now’,‘localtime’)))
对first_name创建唯一索引uniq_idx_firstname，对last_name创建普通索引idx_lastname

create unique index uniq_idx_firstname on actor(first_name);
create index idx_lastname on actor(last_name);

39、针对actor表创建视图actor_name_view

题目描述
针对actor表创建视图actor_name_view，只包含first_name以及last_name两列，并对这两列重新命名，first_name为first_name_v，last_name修改为last_name_v：
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime(‘now’,‘localtime’)))

create view  actor_name_view as 
select first_name as first_name_v,last_name as last_name_v
from actor

40、针对上面的salaries表emp_no字段创建索引idx_emp_no

题目描述
针对salaries表emp_no字段创建索引idx_emp_no，查询emp_no为10005, 使用强制索引。
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 index idx_emp_no on salaries(emp_no);

SELECT * FROM salaries  
INDEXed by idx_emp_no WHERE emp_no = 10005

41、在last_update后面新增加一列名字为create_date

题目描述
存在actor表，包含如下列信息：
CREATE TABLE IF NOT EXISTS actor (
actor_id smallint(5) NOT NULL PRIMARY KEY,
first_name varchar(45) NOT NULL,
last_name varchar(45) NOT NULL,
last_update timestamp NOT NULL DEFAULT (datetime(‘now’,‘localtime’)));
现在在last_update后面新增加一列名字为create_date, 类型为datetime, NOT NULL，默认值为’0000 00:00:00’

alter table actor 
add 'create_date' 
datetime not null default '0000-00-00 00:00:00'

42、构造一个触发器audit_log

题目描述
构造一个触发器audit_log，在向employees_test表中插入一条数据的时候，触发插入相关的数据到audit中。
CREATE TABLE employees_test(
ID INT PRIMARY KEY NOT NULL,
NAME TEXT NOT NULL,
AGE INT NOT NULL,
ADDRESS CHAR(50),
SALARY REAL
);
CREATE TABLE audit(
EMP_no INT NOT NULL,
NAME TEXT NOT NULL
);

CREATE TRIGGER audit_log AFTER INSERT ON employees_test
BEGIN
    INSERT INTO audit VALUES (NEW.ID, NEW.NAME);
END;

53、删除emp_no重复的记录，只保留最小的id对应的记录。

题目描述
删除emp_no重复的记录，只保留最小的id对应的记录。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

insert into titles_test values (‘1’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘2’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘3’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘4’, ‘10004’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘5’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘6’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘7’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’);

delete from titles_test where id not in (
select min(id) from titles_test
group by emp_no)

• 删除重复记录的emp_no，可以分组求最小的min

54、将所有to_date为9999-01-01的全部更新为NULL

题目描述
将所有to_date为9999-01-01的全部更新为NULL,且 from_date更新为2001-01-01。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

insert into titles_test values (‘1’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘2’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘3’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘4’, ‘10004’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘5’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘6’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘7’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’);

update titles_test
set to_date = NULL ,from_date = "2001-01-01"
where to_date = "9999-01-01"

• set 可以为 =null

55、将id=5以及emp_no=10001的行数据替换成id=5以及emp_no=10005

题目描述
将id=5以及emp_no=10001的行数据替换成id=5以及emp_no=10005,其他数据保持不变，使用replace实现。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

insert into titles_test values (‘1’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘2’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘3’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘4’, ‘10004’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘5’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘6’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘7’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’);

update titles_test
set emp_no  = replace(emp_no,10001,10005)
where id = 5

56、将titles_test表名修改为titles_2017

题目描述
将titles_test表名修改为titles_2017。
CREATE TABLE IF NOT EXISTS titles_test (
id int(11) not null primary key,
emp_no int(11) NOT NULL,
title varchar(50) NOT NULL,
from_date date NOT NULL,
to_date date DEFAULT NULL);

insert into titles_test values (‘1’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘2’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘3’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘4’, ‘10004’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’),
(‘5’, ‘10001’, ‘Senior Engineer’, ‘1986-06-26’, ‘9999-01-01’),
(‘6’, ‘10002’, ‘Staff’, ‘1996-08-03’, ‘9999-01-01’),
(‘7’, ‘10003’, ‘Senior Engineer’, ‘1995-12-03’, ‘9999-01-01’);

alter table titles_test
rename to titles_2017

57、在audit表上创建外键约束，其emp_no对应employees_test表的主键id

题目描述
在audit表上创建外键约束，其emp_no对应employees_test表的主键id。
CREATE TABLE employees_test(
ID INT PRIMARY KEY NOT NULL,
NAME TEXT NOT NULL,
AGE INT NOT NULL,
ADDRESS CHAR(50),
SALARY REAL
);

CREATE TABLE audit(
EMP_no INT NOT NULL,
create_date datetime NOT NULL
);

DROP TABLE audit;
CREATE TABLE audit(
    EMP_no INT NOT NULL,
    create_date DATETIME NOT NULL,
    FOREIGN KEY(EMP_no) REFERENCES employees_test(ID));

58、如何获取emp_v和employees有相同的数据no

题目描述
存在如下的视图：
create view emp_v as select * from employees where emp_no >10005;
如何获取emp_v和employees有相同的数据？
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 e1.*
from employees e1,emp_v
where e1.emp_no = emp_v.emp_no

select *
from emp_v

select * from employees where emp_no >10005;

59、将所有获取奖金的员工当前的薪水增加10%

题目描述
将所有获取奖金的员工当前的薪水增加10%。
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
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));

UPDATE salaries
SET salary = salary * 1.1
where to_date = '9999-01-01' and emp_no in (
select emp_no from emp_bonus)

60、针对库中的所有表生成select count(*)对应的SQL语句

题目描述
针对库中的所有表生成select count(*)对应的SQL语句
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 emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
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));
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 "select count(*) from "||name||";" as cnts  from sqlite_master
where type='table'

61、将employees表中的所有员工的last_name和first_name通过(’)连接起来。

题目描述
将employees表中的所有员工的last_name和first_name通过(’)连接起来。
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));
输出格式:

select last_name|| "'" ||first_name
from employees

62、查找字符串’10,A,B’ 中逗号’,'出现的次数cnt

题目描述
查找字符串’10,A,B’ 中逗号’,'出现的次数cnt。

select length('10,A,B')- length(replace('10,A,B',",","")) as cnt

63、获取Employees中的first_name

题目描述
获取Employees中的first_name，查询按照first_name最后两个字母，按照升序进行排列
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 first_name
from employees
order by substr(first_name,-2)

• substr 切割字符串

64、按照dept_no进行汇总

题目描述
按照dept_no进行汇总，属于同一个部门的emp_no按照逗号进行连接，结果给出dept_no以及连接出的结果employees
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));

select dept_no,group_concat(emp_no) as employees
from dept_emp
group by dept_no

• 使用group_concat函数

65、查找排除当前最大、最小salary之后的员工的平均工资avg_salary

题目描述
查找排除当前最大、最小salary之后的员工的平均工资avg_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 avg(salary) as avg_salary
from salaries
where to_date = "9999-01-01"
and salary not in
(select max(salary) from salaries)
and salary not in
(select min(salary) from salaries)

66、分页查询employees表，每5行一页，返回第2页的数据

题目描述
分页查询employees表，每5行一页，返回第2页的数据
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
limit 5,5

select *
from employees
limit 5 offset 5

67、获取所有员工的emp_no

题目描述
获取所有员工的emp_no、部门编号dept_no以及对应的bonus类型btype和received ，没有分配具体的员工不显示
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));
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));
create table emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);

select e.emp_no,de.dept_no,eb.btype,eb.recevied
from employees e inner join dept_emp de
on e.emp_no = de.emp_no left join emp_bonus eb
on de.emp_no = eb.emp_no

68、使用含有关键字exists查找未分配具体部门的员工的所有信息。

题目描述
使用含有关键字exists查找未分配具体部门的员工的所有信息。
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));
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));

select * from employees
where  not exists
(select emp_no
from dept_emp where emp_no = employees.emp_no)

69、获取employees中的行数据，且这些行也存在于emp_v中

题目描述
存在如下的视图：
create view emp_v as select * from employees where emp_no >10005;
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));
获取employees中的行数据，且这些行也存在于emp_v中。注意不能使用intersect关键字。

select e.*
from employees e,emp_v
where e.emp_no = emp_v.emp_no

70、获取有奖金的员工相关信息。

获取有奖金的员工相关信息。
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));
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 emp_bonus(
emp_no int not null,
recevied datetime not null,
btype smallint not null);
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));
给出emp_no、first_name、last_name、奖金类型btype、对应的当前薪水情况salary以及奖金金额bonus。 bonus类型btype为1其奖金为薪水salary的10%，btype为2其奖金为薪水的20%，其他类型均为薪水的30%。 当前薪水表示to_date=‘9999-01-01’
输出格式:

select e.emp_no, e.first_name, e.last_name, eb.btype, s.salary,
(case when eb.btype =  1 then s.salary*0.1
when eb.btype = 2 then s.salary*0.2
else s.salary * 0.3 end) as bonus
from employees e, emp_bonus eb, salaries s
where e.emp_no = eb.emp_no and e.emp_no  = s.emp_no and s.to_date = '9999-01-01'

select e.emp_no,e.first_name,e.last_name,eb.btype,s.salary,
(s.salary*eb.btype*1.0)/10 as bonus
from employees e,emp_bonus eb,salaries s
where e.emp_no = eb.emp_no and e.emp_no = s.emp_no
and s.to_date = '9999-01-01'

71、统计salary的累计和running_total

题目描述
按照salary的累计和running_total，其中running_total为前两个员工的salary累计和，其他以此类推。 具体结果如下Demo展示。。
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 s1.emp_no,s1.salary,sum(s2.salary)
from salaries s1,salaries s2
where s1.to_date = "9999-01-01"
and s2.to_date = "9999-01-01"
and s2.emp_no <= s1.emp_no
group by s1.emp_no

SELECT emp_no,salary,
SUM(salary) OVER (ORDER BY emp_no) AS running_total
FROM salaries
WHERE to_date = '9999-01-01';

• SUM开窗函数,按列累加

72、对于employees表中，给出奇数行的first_name

题目描述
对于employees表中，给出奇数行的first_name
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 e1.first_name
from employees e1
where (select count(*)
      from employees e2
      where e2.first_name <= e1.first_name) %2=1

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