剑指offer SQL训练

其他 2019-03-12 03:30:37 阅读次数: 0
版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/wstcjf/article/details/77997291

1.  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`));
答案： 
select *
from employees as e
order by hire_date desc
limit 0,1  --limit后参数为开始点，及长度

2. 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`));
答案： 
select *
from employees as e
order by hire_date desc 
limit 2,1

3. 

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 
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`));

答案：

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

4. 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`));
答案： 
select e.last_name,e.first_name,d.dept_no
from dept_emp as d inner join employees as e
on d.emp_no = e.emp_no

5. 查找所有员工的last_name和first_name以及对应部门

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`));
答案： 
select e.last_name,e.first_name,d.dept_no
from employees as e left join dept_emp as d
on e.emp_no = d.emp_no

6. 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`));
答案： 
select e.emp_no,s.salary
from employees as e inner join salaries as s
on e.emp_no = s.emp_no and e.hire_date = s.from_date
order by s.emp_no desc

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

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`));
答案： 
select emp_no,count(emp_no) as t
from salaries 
group by emp_no
having t > 15

8. 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`));
答案： 
select distinct s.salary
from salaries as s
where to_date = '9999-01-01'
order by s.salary desc

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'

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`));
答案： 
select d.dept_no,d.emp_no,s.salary
from salaries as s inner join dept_manager as d
on s.emp_no = d.emp_no and s.to_date = '9999-01-01' and s.to_date = d.to_date

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`));

答案：

select e.emp_no
from employees as e
where e.emp_no not in (select d.emp_no from dept_manager as d )
select e.emp_no
from employees as e
except 
select d.emp_no 
from dept_manager as d

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`));

答案：

select e.emp_no,m.emp_no
from dept_emp as e inner join dept_manager as m
on e.dept_no = m.dept_no and e.to_date = '9999-01-01' and e.to_date = m.to_date and e.emp_no != m.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`));

答案：

select d.dept_no,d.emp_no,max(s.salary)
from dept_emp as d inner join salaries as s 
on d.emp_no = s.emp_no and d.to_date = s.to_date and d.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);

答案：

select titles.title,count(title) 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);

答案：

select titles.title,count(distinct titles.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`));

答案：

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

REATE 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);


答案：

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

答案：

select emp_no,salary
from salaries 
where salary = 
(select salary 
 from salaries 
 group by salary 
 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`));

答案：

select e.emp_no, max(s.salary) as salary, e.last_name, e.first_name 
from employees as e inner join salaries as s 
on e.emp_no = s.emp_no and s.to_date = '9999-01-01'
where 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`));


答案：

select e.last_name,e.first_name,d2.dept_name
from (employees as e left join dept_emp as d 
on e.emp_no = d.emp_no) as d1 left join departments as d2
on d1.dept_no = d2.dept_no


20. 查找员工编号emp_now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`));

答案：

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

21. 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`));


答案：

select a.emp_no,(a.salary-b.salary) as growth
from (select e.emp_no,s.salary from employees as e inner join salaries as s 
on e.emp_no = s.emp_no and s.to_date = '9999-01-01') as a  --生成员工当前薪资表
inner join
(select e1.emp_no,s1.salary from employees as e1 inner join salaries as s1
 on e1.emp_no = s1.emp_no and e1.hire_date = s1.from_date) as b  --生成员工入职薪资表
on a.emp_no = b.emp_no
order by growth

22. 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`));


答案： 

select a.dept_no,a.dept_name,count(s.salary) as sum
from departments as a inner join dept_emp as b on a.dept_no = b.dept_no
inner join salaries as s on b.emp_no = s.emp_no
group by a.dept_no

23. 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`));

答案： 

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

24. 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`));


答案： 

select d.dept_no,d.emp_no,s.salary
from dept_emp as d inner join salaries as s
on d.emp_no == s.emp_no and d.to_date = s.to_date and d.to_date='9999-01-01'
where d.emp_no not in (select emp_no from dept_manager)

25. 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`));
答案：

链接：https://www.nowcoder.com/questionTerminal/f858d74a030e48da8e0f69e21be63bef
来源：牛客网
本题主要思想是创建两张表（一张记录当前所有员工的工资，另一张只记录部门经理的工资）进行比较，具体思路如下：
1、先用INNER JOIN连接salaries和demp_emp，建立当前所有员工的工资记录sem
2、再用INNER JOIN连接salaries和demp_manager，建立当前所有员工的工资记录sdm
3、最后用限制条件sem.dept_no = sdm.dept_no AND sem.salary > sdm.salary找出同一部门中工资比经理高的员工，并根据题意依次输出emp_no、manager_no、emp_salary、manager_salary 

select aa.emp_no,bb.emp_no,aa.salary,bb.salary
from (dept_emp as a inner join salaries as s
on a.emp_no = s.emp_no and s.to_date = '9999-01-01') as aa inner join 
(dept_manager as b inner join salaries as s1
on b.emp_no = s1.emp_no and s1.to_date = '9999-01-01' ) as bb
on aa.dept_no = bb.dept_no
where aa.salary > bb.salary

26. 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);

答案： 
select d.dept_no,d.dept_name,t.title,count(t.title) as count
from departments d inner join dept_emp e on d.dept_no == e.dept_no
inner join titles t on e.emp_no = t.emp_no and t.to_date = e.to_date and t.to_date = '9999-01-01'
group by d.dept_no,t.title

27. 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`));


答案： 

select s1.emp_no,s1.from_date,(s1.salary - s2.salary)as salary_growth
from salaries as s1,salaries as s2
where s1.emp_no = s2.emp_no 
and salary_growth > 5000 
and (strftime('%Y', s1.to_date) - strftime('%Y', s2.to_date) = 1 
or strftime('%Y',s1.from_date) - strftime('%Y',s2.from_date) =1)
order by salary_growth desc

28. 查找描述信息中包括robot的电影对应的分类(分类原先对应电影数量>=5)名称以及电影数目

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);

答案： 
select name,count(name)
from film,film_category,category
where film.description like '%robot%' and 
film.film_id= film_category.film_id and 
film_category.category_id= category.category_id and 
category.category_id 
in (select category_id from film_category group by category_id having count(film_id)>=5)


28. 使用join查询方式找出没有分类的电影id以及名称

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);

答案： 
select f.film_id,f.title
from film as f left join film_category as fc
on f.film_id = fc.film_id
where fc.category_id is NULL


29. 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

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

30. 获取select * from employees对应的执行计划

答案： 
explain select * from employees

31. 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`));

答案： 
select (last_name || ' ' || first_name) as Name
from employees


32. 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

答案： 
create table 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'))
);

33. 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


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

答案： 
insert or ignore into actor values('3','ED','CHASE','2006-02-15 12:34:33')
35. 创建一个actor_name表，将actor表中的所有first_name以及last_name导入改表

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

答案： 
create table 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

36.  对first_name创建唯一索引uniq_idx_firstname，对last_name创建普通索引idx_lastname

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')))

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

37.  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，fist_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')))

答案： 

create view actor_name_view as 
select first_name as fist_name_v,last_name as last_name_v
from actor
38.   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);

答案： 
select * 
from salaries indexed by idx_emp_no
where emp_no = '10005'

39.  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'

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')));
答案： 
alter table actor add create_date datetime not null default('0000-00-00 00:00:00')

40. 构造一个触发器audit_log，在向employees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
);

答案： 
create trigger audit_log after insert on employees_test
begin
     insert into audit values(new.ID,new.NAME);
end

41. 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);

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

42.  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);

答案： 
update titles_test set to_date = null, from_date = '2001-01-01'
where to_date = '9999-01-01'
43. 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);

答案： 
replace into titles_test values(5,10005,'Senior Engineer','1986-06-26','9999-01-01')

44. 将titles_test表名修改为titles_2017

答案： 
alter table titles_test rename to titles_2017

45. 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
);

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

46. 如何获取emp_v和employees有相同的数据
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`));

答案： 
select * 
from emp_v

47. 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`));

答案： 
update salaries set salary = salary * 1.1
where emp_no in
(select b.emp_no from emp_bonus as b inner join salaries as s 
on b.emp_no = s.emp_no and s.to_date = '9999-01-01'
)


48. 针对库中的所有表生成select count(*)对应的SQL语句

答案： 
SELECT "select count(*) from " || name || ";" AS cnts
FROM sqlite_master WHERE type = 'table'

49. 将employees表中的所有员工的last_name和first_name通过(')连接起来

答案： 
SELECT last_name || "'" || first_name 
FROM employees

50. 查找字符串'10,A,B' 中逗号','出现的次数cnt

答案： 
select length('10,A,B') - length(replace('10,A,B',",",""))
51. 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`));

答案： 
select first_name
from employees
order by substr(first_name,-2)

52. 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`));
答案： 
select dept_no,group_concat(emp_no) as employees
from dept_emp
group by dept_no
53. 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`));

答案： 
select avg(salary) avg_salary
from salaries
where salary not in
(select max(salary) from salaries
 union
 select min(salary) from salaries) 
and to_date = '9999-01-01'
54. 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`));
答案： 
select * from employees
limit 5,5

55.  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`));

答案： 
select * 
from employees ee 
where not exists (
 select emp_no from dept_emp de where de.emp_no = ee.emp_no
)

















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