数据库-刷题

1、题目描述
汇总各个部门当前员工的title类型的分配数目,即结果给出部门编号dept_no、dept_name、其部门下所有的当前(dept_emp.to_date = ‘9999-01-01’)员工的当前(titles.to_date = ‘9999-01-01’)title以及该类型title对应的数目count
(注:因为员工可能有离职,所有dept_emp里面to_date不为’9999-01-01’就已经离职了,不计入统计,而且员工可能有晋升,所以如果titles.to_date 不为 ‘9999-01-01’,那么这个可能是员工之前的职位信息,也不计入统计)

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);

输出描述:

数据库-刷题
实现:

select dp.dept_no,dp.dept_name,t.title, count(t.title)
from departments dp,
dept_emp de,
titles t
where dp.dept_no=de.dept_no and
de.emp_no=t.emp_no
and de.to_date='9999-01-01' and
t.to_date='9999-01-01'
group by dp.dept_no,dp.dept_name ,t.title

注意:就是三个表的内连接inner join 然后进行分组,先按照 dept_no 分组,然后按照 dept_name分组,再按照 title分组 即可实现。

2、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)
(数据保证每个员工的每条薪水记录to_date-from_date=1年,而且同一员工的下一条薪水记录from_data=上一条薪水记录的to_data)

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`));
如:插入
INSERT INTO salaries VALUES(10001,52117,'1986-06-26','1987-06-26');
INSERT INTO salaries VALUES(10001,62102,'1987-06-26','1988-06-25');
INSERT INTO salaries VALUES(10002,72527,'1996-08-03','1997-08-03');
INSERT INTO salaries VALUES(10002,72527,'1997-08-03','1998-08-03');
INSERT INTO salaries VALUES(10002,72527,'1998-08-03','1999-08-03');
INSERT INTO salaries VALUES(10003,43616,'1996-12-02','1997-12-02');
INSERT INTO salaries VALUES(10003,43466,'1997-12-02','1998-12-02');

输出描述:
数据库-刷题
SQL代码实现:

select s1.emp_no,s1.from_date, s1.salary-s2.salary as salary_growth
from salaries s1 inner join salaries s2 on
s1.emp_no=s2.emp_no and strftime('%Y',s1.from_date)=strftime('%Y',s2.to_date)
where s1.salary-s2.salary >5000
order by salary_growth DESC

注意:需要保证一年的薪水情况,需要建立两个表格,并进行连接。

3、题目描述

数据库-刷题
数据库-刷题

数据库-刷题
查找描述信息(film.description)中包含robot的电影对应的分类名称(category.name)以及电影数目(count(film.film_id)),而且还需要该分类包含电影总数量(count(film_category.category_id))>=5部

输入为:

INSERT INTO film VALUES(1,'ACADEMY DINOSAUR','A Epic Drama of a Feminist And a Mad Scientist who must Battle a Teacher in The Canadian Rockies');
INSERT INTO film VALUES(2,'ACE GOLDFINGER','A Astounding Epistle of a Database Administrator And a Explorer who must Find a Car in Ancient China');
INSERT INTO film VALUES(3,'ADAPTATION HOLES','A Astounding Reflection of a Lumberjack And a Car who must Sink a Lumberjack in A Baloon Factory');
INSERT INTO film VALUES(4,'AFFAIR PREJUDICE','A Fanciful Documentary of a Frisbee And a Lumberjack who must Chase a Monkey in A Shark Tank');
INSERT INTO film VALUES(5,'AFRICAN EGG','A Fast-Paced Documentary of a Pastry Chef And a Dentist who must Pursue a Forensic Psychologist in The Gulf of Mexico');
INSERT INTO film VALUES(6,'AGENT TRUMAN','A Intrepid Panorama of a robot And a Boy who must Escape a Sumo Wrestler in Ancient China');
INSERT INTO film VALUES(7,'AIRPLANE SIERRA','A Touching Saga of a Hunter And a Butler who must Discover a Butler in A Jet Boat');
INSERT INTO film VALUES(8,'AIRPORT POLLOCK','A Epic Tale of a Moose And a Girl who must Confront a Monkey in Ancient India');
INSERT INTO film VALUES(9,'ALABAMA DEVIL','A Thoughtful Panorama of a Database Administrator And a Mad Scientist who must Outgun a Mad Scientist in A Jet Boat');
INSERT INTO film VALUES(10,'ALADDIN CALENDAR','A Action-Packed Tale of a Man And a Lumberjack who must Reach a Feminist in Ancient China');

INSERT INTO category VALUES(1,'Action','2006-02-14 20:46:27');
INSERT INTO category VALUES(2,'Animation','2006-02-14 20:46:27');
INSERT INTO category VALUES(3,'Children','2006-02-14 20:46:27');
INSERT INTO category VALUES(4,'Classics','2006-02-14 20:46:27');
INSERT INTO category VALUES(5,'Comedy','2006-02-14 20:46:27');
INSERT INTO category VALUES(6,'Documentary','2006-02-14 20:46:27');
INSERT INTO category VALUES(7,'Drama','2006-02-14 20:46:27');
INSERT INTO category VALUES(8,'Family','2006-02-14 20:46:27');
INSERT INTO category VALUES(9,'Foreign','2006-02-14 20:46:27');
INSERT INTO category VALUES(10,'Games','2006-02-14 20:46:27');
INSERT INTO category VALUES(11,'Horror','2006-02-14 20:46:27');
INSERT INTO category VALUES(12,'Music','2006-02-14 20:46:27');
INSERT INTO category VALUES(13,'New','2006-02-14 20:46:27');
INSERT INTO category VALUES(14,'Sci-Fi','2006-02-14 20:46:27');
INSERT INTO category VALUES(15,'Sports','2006-02-14 20:46:27');
INSERT INTO category VALUES(16,'Travel','2006-02-14 20:46:27');

INSERT INTO film_category VALUES(1,6,'2006-02-14 21:07:09');
INSERT INTO film_category VALUES(2,11,'2006-02-14 21:07:09');
INSERT INTO film_category VALUES(3,6,'2006-02-14 21:07:09');
INSERT INTO film_category VALUES(4,11,'2006-02-14 21:07:09');
INSERT INTO film_category VALUES(5,6,'2006-02-14 21:07:09');
INSERT INTO film_category VALUES(6,6,'2006-02-14 21:07:09');
INSERT INTO film_category VALUES(7,5,'2006-02-14 21:07:09');
INSERT INTO film_category VALUES(8,6,'2006-02-14 21:07:09');
INSERT INTO film_category VALUES(9,11,'2006-02-14 21:07:09');
INSERT INTO film_category VALUES(10,15,'2006-02-14 21:07:09');

输出为:

数据库-刷题
代码实现:

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

注意:需要找出电影分类表中,电影类别的个数大于5的表格,然后需要将三个表格进行内连接。

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数据库-刷题

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