数据库SQL实战练习4

1、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 d.dept_no , e.emp_no , s.salary
from dept_emp d , employees e , salaries s
where d.emp_no = e.emp_no
and s.emp_no = e.emp_no
and d.to_date = '9999-01-01'
and s.to_date = '9999-01-01'
and e.emp_no not in (select emp_no from dept_manager)

2、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 EmpInfo.emp_no as emp_no, ManagerInfo.emp_no as manager_no , EmpInfo.salary as emp_salary, ManagerInfo.salary as manager_salary
from (select de.emp_no , s.salary , de.dept_no
        from dept_emp de , salaries s , dept_manager dp
        where de.emp_no = s.emp_no
        and dp.dept_no = de.dept_no
        and s.to_date = '9999-01-01'
        and de.emp_no not in (select emp_no from dept_manager)) EmpInfo , (select de.emp_no , sa.salary , de.dept_no
                                                                            from dept_manager de , salaries sa
                                                                            where sa.to_date = '9999-01-01'
                                                                            and de.emp_no = sa.emp_no) as ManagerInfo
where  EmpInfo.dept_no = ManagerInfo.dept_no
and EmpInfo.salary > ManagerInfo.salary

3、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 de.dept_no, dp.dept_name, t.title, COUNT(t.title) AS count
FROM titles AS t INNER JOIN dept_emp AS de
ON t.emp_no = de.emp_no AND de.to_date = '9999-01-01' AND t.to_date = '9999-01-01'
INNER JOIN departments AS dp
ON de.dept_no = dp.dept_no
GROUP BY de.dept_no, t.title
*/

SELECT c.dept_no, dp.dept_name, c.title, COUNT(c.title) AS count
FROM (select de.dept_no ,t.title
      from titles AS t,dept_emp AS de 
       where t.emp_no = de.emp_no AND de.to_date = '9999-01-01' AND t.to_date = '9999-01-01') as c , departments AS dp 
where c.dept_no = dp.dept_no
GROUP BY dp.dept_no, c.title

4、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));

-- (strftime('%Y',s2.to_date) - strftime('%Y',s1.to_date) = 1
-- or strftime('%Y',s2.from_date) - strftime('%Y',s1.from_date) =  1 
-- 一年可能涨两次薪水 也可能两年涨一次薪水
-- strftime('%Y',s2.to_date) - strftime('%Y',s1.from_date) = 1

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

5、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

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);

last_update	电影id和分类id对应关系的最后更新时间
category_id	电影分类id
last_update	电影分类最后更新时间
description	电影描述信息

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

--  先查询出电影数量大于5的分类的集合，然后限定category必须在该集合中

6、使用join查询方式找出没有分类的电影id以及名称（见第五题表）

-- 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 ca
on ca.category_id = fc.category_id
where f.film_id not in (select film_id from film_category)
-- 没有分类用 is null就可以了  但是为什么我写 not in (select category_id from category )

7、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（见第五题表）

-- 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 film f, film_category ca
where f.film_id = ca.film_id
and ca.category_id in (select category_id 
                        from category
                        where name = 'Action')

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