牛客数据库SQL实战题(21-30题)
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 now.emp_no, (now.salary-pre.salary) as growth
from (select em.emp_no, sa.salary
from employees as em
join salaries as sa
on em.emp_no=sa.emp_no
where sa.to_date='9999-01-01') as now
join (select em.emp_no, sa.salary
from employees as em
join salaries as sa
on em.emp_no=sa.emp_no
where em.hire_date=sa.from_date) as pre
on now.emp_no=pre.emp_no
order by growth;
分别建两张表，一张表是emp_no和对应的开始工资，一张表是emp_no和对应的当前工资。然后将两张表连接起来，做减法。
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 de.dept_no, dename.dept_name, count(sa.salary) as sum
from (dept_emp as de
join departments as dename on dename.dept_no=de.dept_no)
join salaries as sa on de.emp_no=sa.emp_no
group by de.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 sa1.emp_no, sa1.salary, count(distinct sa2.salary) as rank
from salaries as sa1
left join salaries as sa2
on sa2.salary>=sa1.salary
where sa1.to_date='9999-01-01' and sa2.to_date='9999-01-01'
group by sa1.emp_no
order by sa1.salary desc, sa1.emp_no;
排序这个字段需要造出一个字段并count，为了构造出rank，我们可以复用salaries表，对salaries1表中的每一个salary，对大于或等于s1.salary的sa.salary进行count。
参考牛客网，在支持ROW_NUMBER、RANK、DENSE_RANK等函数的SQL Server数据库中，有以下参考代码，可惜在本题的SQLite数据库中不支持。
SELECT emp_no, salaries, DENSE_RANK() OVER(ORDER BY salary DESC) AS rank
WHERE to_date ='9999-01-01'
ORDER BY salary DESC, 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 dema.dept_no, em.emp_no, sa.salary
from employees as em
join dept_emp as de
on em.emp_no=de.emp_no
join dept_manager as dema
on de.dept_no=dema.dept_no
join salaries as sa
on em.emp_no=sa.emp_no
where sa.to_date='9999-01-01' and em.emp_no<>dema.emp_no
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，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
select em.emp_no, ma.emp_no as manager_no, sa1.salary as emp_salary, sa2.salary as manager_salary
from dept_emp as em
join dept_manager as ma
on em.dept_no=ma.dept_no
and em.to_date='9999-01-01'
and ma.to_date='9999-01-01'
and em.emp_no<>ma.emp_no
join salaries as sa1
on em.emp_no=sa1.emp_no and sa1.to_date='9999-01-01'
join salaries as sa2
on ma.emp_no=sa2.emp_no and sa2.to_date='9999-01-01'
where sa1.salary>sa2.salary
重复利用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 de.dept_no, name.dept_name, ti.title, count(ti.title)
from dept_emp as de
join departments as name
on de.dept_no=name.dept_no and de.to_date='9999-01-01'
join titles as ti
on de.emp_no=ti.emp_no and ti.to_date='9999-01-01'
group by de.dept_no, ti.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 s2.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 salary_growth > 5000
and (strftime("%Y",s2.to_date) - strftime("%Y",s1.to_date) = 1
or strftime("%Y",s2.from_date) - strftime("%Y",s1.from_date) = 1 )
order by salary_growth desc
每年薪水涨幅的定义并不是非常明确。
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);
答案
select c.name, count(fc.film_id)
from (select category_id, count(film_id) as amount
from film_category
group by category_id
having amount>=5) as cc,
film as f, category as c, film_category as fc
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;
首先，此题所说的该分类对应电影数量>=5，是该分类下所有电影的数量大>=5，而不是筛选了描述中有robot字段之后电影数量依然>=5的分类。
其次，这题的代码还是让人摸不着头脑，第一行的select c.name, count(fc.film_id)中的count(fc.film_id)很奇怪，还没有group by。用join各表的方法写的代码无法通过。作者才疏学浅，如果你有好的想法请联系我。
29、使用join查询方式找出没有分类的电影id以及名称
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);
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;
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
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);
答案
子查询方式：
select f.title, f.description
from film as f
where f.film_id
in (select fc.film_id
from film_category as fc
where fc.category_id
in (select c.category_id
from category as c
where name='Action'))
非子查询方式：
select f.title,f.description
from film as f
inner join film_category as fc
on f.film_id = fc.film_id
inner join category as c
on c.category_id = fc.category_id
where c.name = 'Action';
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