SQL实战

Posted 2020-11-02 箬笠蓑衣

一、

 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_noy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`));

employees 中tem_no和salaries中tem_no插入，然后找最大最小，求值

1、

select sCurrent.emp_no,(sCurrent.salary-sStart.salary) as growth
from (select e.emp_no,s.salary from employees as e,salaries as s where e.emp_no=s.emp_no and s.to_date="9999-01-01") as sCurrent  --找到当前薪水
join (select e.emp_no,s.salary from employees as e,salaries as s where e.emp_no=s.emp_no and s.from_date=e.hire_date) as sStart  --找到初始入职薪水
on sCurrent.emp_no=sStart.emp_no   --合并表，相减
order by growth 

生成新表，两个新表连接后再减，而不能在一个表里多次跨行减。　　

2、

更好理解

select a.emp_no,(b.salary-c.salary) as growth
from employees as a  --往该表添加
join salaries as b on a.emp_no=b.emp_no and b.to_date="9999-01-01"   --加入当前薪水
join salaries as c on a.emp_no=c.emp_no and c.from_date=a.hire_date  --加入最初薪水
order by growth

 

二、

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`));

 

1、departments 连接dept_emp ，然后再连接 salaries ，group by 分组，统计个数

select a.dept_no,a.dept_name,count(a.dept_no) 

from departments as a 

join dept_emp as b on a.dept_no=b.dept_no

join salaries as c on b.emp_no=c.emp_no

group by a.dept_no

2、思路一样，换用from where 表示

select d.dept_no,d.dept_name,count(d.dept_no) as sum

from (select * from departments as a,dept_emp as b where a.dept_no=b.dept_no)as d,salaries as c 

where d.emp_no=c.emp_no

group by d.dept_no

 

三、

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`));

1、题有点难度

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 ASC

思路如下

学会where语句中判断语句 的应用

两个条件下排序的应用。

 

三、

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`));

11个人6个部门，10008和10011离职。

从dept_manager中找出部门对应的经理，从dept_emp中找出部门对应的员工，从salaries找出薪资，关键是如何从员工中剔除经理。

1、

SELECT de.dept_no, s.emp_no, s.salary 

FROM (employees AS e INNER JOIN salaries AS s ON s.emp_no = e.emp_no AND s.to_date = \'9999-01-01\')--每个人的当前薪资

INNER JOIN dept_emp AS de ON e.emp_no = de.emp_no --所有员工待的部门

WHERE de.emp_no NOT IN (SELECT emp_no FROM dept_manager WHERE to_date = \'9999-01-01\') --构建经理的集合，not in 集合的输出

from...join...as A 。这里A是连接后表的名称。

not in 后一定跟着一个表而非一个数。

2、不用employees表

select a.dept_no,b.emp_no,b.salary

from dept_emp as a join salaries as b 

on a.emp_no=b.emp_no and b.to_date="9999-01-01"  --员工当前薪水

where a.emp_no not in (select emp_no from dept_manager where dept_manager.to_date="9999-01-01")

3、

select de.dept_no,de.emp_no,(select salary from salaries where emp_no=de.emp_no and to_date=\'9999-01-01\') as salary 
from dept_emp de left join dept_manager dm 
on de.emp_no=dm.emp_no 
where de.to_date=\'9999-01-01\' and dm.emp_no is null;

利用左连接求差集，没有经理的会出现null 。由此判断　

还有一点，是　(select salary from salaries where emp_no=de.emp_no and to_date=\'9999-01-01\')，输出位置这样也可以。

 

四、

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`));

　

具体数据见上题

1、

select a.emp_no as emp_no,b.emp_no as manager_no,a.salary as emp_salary,b.salary as manager_salary

from (dept_emp as de join salaries as s on de.emp_no=s.emp_no and s.to_date="9999-01-01") as a --所有员工薪资表a,

join (dept_manager as dp join salaries as s on dp.emp_no=s.emp_no and s.to_date="9999-01-01") as b --经理薪资表 b,

on de.dept_no=dp.dept_no and dp.to_date="9999-01-01" and a.salary>b.salary --a和b 按条件连接

　

五、

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);

1、首先department 和dept_emp 用join连接，可以得到部门里的员工，

再和title join连接，得到每个员工的职称

按照部门分类，再按职称分类，统计各个职称的个数

核心是group by可以按照顺序对两个条件分类的使用。

select a.dept_no,a.dept_name,t.title,count(t.title)

from (departments as dp join dept_emp as de on de.dept_no=dp.dept_no and de.to_date="9999-01-01") as a

join titles as t on a.emp_no=t.emp_no and t.to_date="9999-01-01" 

group by a.dept_no,t.title

 

六、

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,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.from_date)-strftime(\'%Y\',s2.from_date)=1) or (strftime(\'%Y\',s1.to_date)-strftime(\'%Y\',s2.to_date)=1))--and中嵌套or的条件判断

order by salary_growth desc

 

七、

1、

注意该分类电影数量》=5，是所有的电影，而不是仅仅只包含robot的电影

思路：找到robot的电影，找到电影的分类，找到该分类的名称，比较该分类是否在大于5的范围内。

正则表达式来判断robot的存在。

select name,count(name)
from film,film_category,category
where film.description like \'%robot%\' --找到含有robot 电影
and film.film_id= film_category.film_id and --找到电影的分类
film_category.category_id= category.category_id --找到分类名
and category.category_id in 
(select category_id from film_category group by category_id having count(film_id)>=5)--分类大于5，不考虑是否有robot

join 方式

select ca.name,count(ca.name)
from film as fi join film_category as fc
on fi.description like "%robot%" and fi.film_id=fc.film_id
join category as ca 
on ca.category_id=fc.category_id 
where fc.category_id in (select category_id from film_category group by category_id having count(category_id)>=5)

 

八、

使用join查询方式找出没有分类的电影id以及名称

1、左连接 取差

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

2、

不用连接的方式

找出film_category中film_id的集合，film中的film_id没有在集合中的就输出

select f.film_id,f.title from film as f
where f.film_id not in (select film_id from film_category)

　　

九、

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

1、

非子查询的方法，join

按着电影名、电影类型将三个表连接起来，然后限制Action条件就可以了。

select fi.title,fi.description

from film_category as fc join film as fi on fc.film_id=fi.film_id 

join category as ca on fc.category_id=ca.category_id and ca.name="Action"

2、子查询

层层嵌套利用where查询

select fi.title,fi.description from film as fi 
  where fi.film_id in
    (select fc.film_id from film_category as fc
        where fc.category_id in 
           (select ca.category_id from category as ca where ca.name="Action"))

 

十、

题目描述

获取select * from employees对应的执行计划

 

explain select * from employees