先根据考察的知识点对题目做个整理:
聚集函数:
1.查找最晚入职员工的所有信息
order by(limit):
2.查找入职员工时间排名倒数第三的员工所有信息
6.查找所有员工入职时候的薪水情况
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
内连接(inner join)
3.查找当前薪水详情以及部门编号dept_no
4.查找所有已经分配部门的员工的last_name和first_name
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
24.获取所有非manager员工当前的薪水情况
左连接(left join)
5.查找所有员工的last_name和first_name以及对应部门编号dept_no
11.获取所有员工当前的manager
19.查找所有员工的last_name和first_name以及对应的dept_name
自连接
21.查找所有员工自入职以来的薪水涨幅情况dept_name
group by
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
12.获取所有部门中当前员工薪水最高的相关信息
13.从titles表获取按照title进行分组
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
15.查找employees表
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
22.统计各个部门对应员工涨幅的次数总和
26.汇总各个部门当前员工的title类型的分配数目
distinct
8.找出所有员工当前薪水salary情况
子查询
10.获取所有非manager的员工emp_no
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
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
25.获取员工其当前的薪水比其manager当前薪水还高的相关信息
多个考点(或者说我不会的点。。。)
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
21.查找所有员工自入职以来的薪水涨幅情况dept_name
23.对所有员工的薪水按照salary进行按照1-N的排名
25.获取员工其当前的薪水比其manager当前薪水还高的相关信息 (太难了,我太难了,完全没有思路)
27.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:
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
));
代码:
SELECT *
FROM employees
WHERE hire_date=(SELECT max(hire_date) FROM employees)
题目描述:
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
));
代码:
select *
from employees
order by hire_date
desc limit 2,1
/*select * from tableName limit i,n
# tableName:表名
# i:为查询结果的索引值(默认从0开始),当i=0时可省略i
# n:为查询结果返回的数量
# i与n之间使用英文逗号","隔开 */
/*本题直接按入职时间降序排列,然后取第三个*/
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
));
Remark:我这道题都没看清题目,写起来落这落那的,补了半天条件才通过,被自己菜哭。当初学的数据库和SQL都还老师了o(╥﹏╥)o
代码:
select salaries.emp_no,salaries.salary,salaries.from_date,salaries.to_date,dept_manager.dept_no
from salaries
inner join dept_manager
on salaries.emp_no=dept_manager.emp_no
and salaries.to_date='9999-01-01'
and dept_manager.to_date='9999-01-01'
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
));
代码:
select employees.last_name,
employees.first_name,
dept_emp.dept_no
from employees
inner join dept_emp
on employees.emp_no=dept_emp.emp_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
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
));
代码:
select a.last_name,a.first_name,b.dept_no
from employees a
left join dept_emp b
#此处不论有没有分配部门的员工都要选出来,所以用左连接
on a.emp_no=b.emp_no
LEFT JOIN 关键字会从左表 (table_name1) 那里返回所有的行,即使在右表 (table_name2) 中没有匹配的行。
具体语法看这里
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
));
Remak:还是没看清题目,写起来丢三落四,被自己菜哭
代码:
select e.emp_no,s.salary
from employees e
left join salaries s
on e.emp_no=s.emp_no
and s.from_date=e.hire_date
order by e.emp_no
desc
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
));
代码:
select *
from(
select emp_no,count(*) t
from salaries
group by emp_no
)
where t>15
select emp_no,count(salary) as t
from salaries
group by emp_no
having count(salary)>15
group by语法
GROUP BY 语句用于结合合计函数,根据一个或多个列对结果集进行分组。
这里根据emp_no进行分组并计算列数,也就是工资涨幅的次数,然后把选出的emp_no和t作为新的表被外层的select查询
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
));
代码:
select distinct salary
from salaries
where to_date='9999-01-01'
order by salary
desc
题目描述:
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
));
代码:
select d.dept_no,d.emp_no,s.salary
from dept_manager as d
inner join salaries as s
on d.emp_no=s.emp_no
and d.to_date='9999-01-01'
and s.to_date='9999-01-01'
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
));
代码:
select e.emp_no
from employees e
left join dept_manager d
on e.emp_no=d.emp_no
where d.emp_no is null
select emp_no
from employees
where emp_no not in (select emp_no from dept_manager)
Remark:最后的条件要用‘where’不能用‘and’。。。
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
));
代码:
select a.emp_no,b.emp_no manager_no
from dept_emp a
inner join dept_manager b
on a.dept_no=b.dept_no
where a.to_date='9999-01-01'
and b.to_date='9999-01-01'
and a.emp_no<>b.emp_no
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
));
代码:
select d.dept_no,d.emp_no,max(s.salary) 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
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);
代码:
select title,count(*) t
from titles
group by title
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);
代码:
select title,count(distinct emp_no) t
from titles
group by title
having t>=2;
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
));
代码:
select *
from employees
where emp_no%2<>0
and last_name <> 'Mary'
order by hire_date
desc
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);
代码:
select t.title,avg(s.salary) as avg
from titles t
inner join salaries s
on t.emp_no=s.emp_no
where t.to_date='9999-01-01'
and s.to_date='9999-01-01'
group by title
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
));
代码:
select emp_no,salary
from salaries
where to_date='9999-01-01'
order by salary
desc
limit 1,1
select emp_no,max(salary) as salary
from salaries
where to_date='9999-01-01'
and salary not in (select max(salary) from salaries);
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
));
代码:
select e.emp_no,max(s.salary) 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'
and s.salary not in
(select max(salary) from salaries where to_date='9999-01-01')
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
));
代码:
select e.last_name,e.first_name,dp.dept_name
from employees e
left join dept_emp d
on e.emp_no=d.emp_no
left join departments dp
on d.dept_no=dp.dept_no
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
));
思路:
1、先分别找到emp_no=10001的员工的第一次工资记录与最后一次工资记录
2、再将最后一次工资记录减去第一次工资记录得到入职以来salary的涨幅,最后用别名growth代替
代码:
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
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
));
代码:
select e.emp_no,(a.salary - b.salary) growth
from
employees e
inner join salaries a
on e.emp_no=a.emp_no and a.to_date='9999-01-01'
inner join salaries b
on e.emp_no=b.emp_no and b.from_date=e.hire_date
order by growth asc
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
));
代码:
select d.dept_no,dp.dept_name,count(*) sum
from dept_emp d
inner join
salaries s
on d.emp_no=s.emp_no
inner join
departments dp
on dp.dept_no=d.dept_no
group by d.dept_no
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
));
代码:
select s1.emp_no,s1.salary,count(distinct s2.salary) 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 rank asc
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
));
代码:
select d.dept_no,d.emp_no,s.salary
from dept_emp d,salaries s
where d.emp_no=s.emp_no
and s.to_date='9999-01-01'
and d.to_date='9999-01-01'
and d.emp_no not in
(select emp_no from dept_manager)
题目描述: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
));
代码:
select se.emp_no emp_no,sm.emp_no as manager_no,
se.salary as emp_salary,sm.salary as manager_salary
from
(
select d.emp_no,d.dept_no,s.salary
from dept_emp d
inner join salaries s
on d.emp_no=s.emp_no
where s.to_date='9999-01-01'
)as se
inner join
(
select dp.emp_no,dp.dept_no,s.salary
from dept_manager dp
inner join salaries s
on dp.emp_no=s.emp_no
where s.to_date='9999-01-01'
) as sm
on se.dept_no=sm.dept_no
where se.salary>sm.salary
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);
代码:
select d.dept_no,dp.dept_name,t.title,count(t.emp_no) as count
from dept_emp d
inner join departments dp
on d.dept_no=dp.dept_no
inner join titles t
on d.emp_no=t.emp_no
and d.to_date='9999-01-01'
and t.to_date='9999-01-01'
group by d.dept_no,t.title
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
));
代码:
select b.emp_no,b.from_date,b.salary-a.salary as salary_growth
from salaries a,salaries b
where a.emp_no=b.emp_no
and b.salary-a.salary>5000
and strftime('%Y',b.to_date)-strftime('%Y',a.to_date)=1
order by salary_growth desc;
28.查找描述信息中包括robot的电影对应的分类名称以及电影数目,而且还需要该分类对应电影
题目描述:
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));
CREATE TABLE category (
category_id tinyint(3) NOT NULL ,
name varchar(25) NOT NULL, last_update
timestamp,
PRIMARY KEY ( category_id ));
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
代码:
select c.name,count(f.film_id)
from
(
select category_id,count(film_id) from film_category
group by category_id having count(film_id)>=5
) as cc,
film f,film_category fc,category c
where f.description like '%robot%'
and f.film_id=fc.film_id
and fc.category_id=c.category_id
and c.category_id=cc.category_id
29.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:
使用join查询方式找出没有分类的电影id以及名称
代码:
select f.film_id,f.title
from film f
left join film_category fc
on f.film_id=fc.film_id
where fc.category_id is null
30.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:
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
代码:
select f.title,f.description
from film f
where f.film_id in
(
select fc.film_id
from film_category fc
where fc.category_id in
(
select c.category_id from category c
where c.name='Action'
)
);
31.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
代码:
explain select * from employees;
explain的用法
32.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
));
代码:
select last_name ||' '|| first_name as Name from employees
Remark:Mysql 数据库可以使用CONCAT或者CONCAT_WS两种函数进行拼接,但是SQLLite得使用||进行拼接
33.创建一个actor表,包含如下列信息
题目描述:
代码:
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'))
);
insert into actor
values
(
1,'PENELOPE','GUINESS','2006-02-15 12:34:33'
),
(
2,'NICK','WAHLBERG','2006-02-15 12:34:33'
)
;
35.批量插入数据,不使用replace操作
题目描述:
代码:
insert or ignore into actor
values
(
'3','ED','CHASE','2006-02-15 12:34:33'
);
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;
37.对first_name创建唯一索引uniq_idx_firstname
题目描述:
代码:
create unique index uniq_idx_firstname on actor(first_name);
create index idx_lastname on actor(last_name);
38.针对actor表创建视图actor_name_view
题目描述:
代码:
create view actor_name_view as
select first_name as first_name_v,last_name as last_name_v
from actor
39.针对上面的salaries表emp_no字段创建索引idx_emp_no
题目描述:
代码:
select * from salaries
indexed by idx_emp_no
where emp_no='10005'
40.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’
代码:
alter table actor
add column create_date datetime NOT NULL default '0000-00-00 00:00:00'
41.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
);
代码:
在这里插入代码片
42.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’);
代码:
delete from titles_test
where id not in (select min(id) from titles_test group by emp_no);
43.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’);
代码:
update titles_test
set to_date=null,from_date='2001-01-01'
where to_date='9999-01-01';
44.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’);
代码:
replace into titles_test
values(5,10005,'Senior Engineer', '1986-06-26', '9999-01-01');
45.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’);
代码:
alter table titles_test
rename to titles_2017;
46.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
);
代码:
drop table audit;
create table audit(
emp_no int not null,
create_date datetime not null,
foreign key(emp_no) references employees_test(ID));
47.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
));
输出格式:
代码:
select * from emp_v
48.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
));
代码:
update salaries
set salary=salary*1.1
where emp_no in (select emp_no from emp_bonus);
49.针对库中的所有表生成select count(*)对应的SQL语句
题目描述:
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
));
输出格式:
代码:
select "select count(*) from "|| name ||";" as cnts
from sqlite_master where type='table'
50.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:
代码:
select last_name ||"'"|| first_name as name from employees;
51.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。
代码:
select length('10,A,B')-length(replace('10,A,B',',','')) as cnt;
52.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
));
输出格式:
代码:
select first_name
from employees
order by substr(first_name,length(first_name)-1);
substr的用法
53.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
));
输出格式:
代码:
select dept_no,group_concat(emp_no) as employees
from dept_emp
group by dept_no;
54.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
));
输出格式:
代码:
select avg(salary) as avg_salary
from salaries
where salary <> (select max(salary) from salaries)
and salary<>(select min(salary) from salaries)
and to_date='9999-01-01';
55.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
));
代码:
select * from employees limit 5,5;
56.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);
输出格式:
代码:
select e.emp_no,d.dept_no,eb.btype,eb.recevied
from employees as e
inner join dept_emp as d
on e.emp_no=d.emp_no
left join emp_bonus as eb
on d.emp_no=eb.emp_no;
57.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
));
输出格式:
代码:
select * from employees
where not exists
(select emp_no from dept_emp
where emp_no=employees.emp_no);
58.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。
输出格式:
代码:
select * from employees where emp_no >10005;
59.获取有奖金的员工相关信息。
题目描述:
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:
代码:
select e.emp_no,e.first_name,e.last_name,eb.btype,s.salary,
(case eb.btype
when 1 then s.salary*0.1
when 2 then s.salary*0.2
else s.salary*0.3 end) as bonus
from employees e
inner join emp_bonus eb
on e.emp_no=eb.emp_no
inner join salaries s
on s.emp_no=e.emp_no
and s.to_date='9999-01-01';
60.统计salary的累计和running_total
题目描述: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
));
输出格式:
代码:
select s1.emp_no,s1.salary,
(select sum(s2.salary) from salaries s2
where s2.emp_no<=s1.emp_no and s2.to_date='9999-01-01') as running_total
from salaries s1
where s1.to_date='9999-01-01'
order by s1.emp_no;
61.对于employees表中,给出奇数行的first_name
题目描述: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
));
输出格式:
代码:
select e1.first_name
from employees e1
where (select count(*) from employees e2
where e2.first_name<=e1.first_name)%2=1