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山东大学 2020级数据库系统 实验五

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What’s more

山东大学 2020级数据库系统 实验一
山东大学 2020级数据库系统 实验二
山东大学 2020级数据库系统 实验三
山东大学 2020级数据库系统 实验四
山东大学 2020级数据库系统 实验五
山东大学 2020级数据库系统 实验六
山东大学 2020级数据库系统 实验七
山东大学 2020级数据库系统 实验八、九

写在前面

做数据库实验一定要静得下心来，才能发现其中的错误然后进行改正。同时，如果发现 SQL 语句总是报错，“一定是你错了，只是不知道错在哪里！”

其次，SQL 语句中较为复杂的点博主都进行了注释，希望大家一定要看懂思路后自己写一遍，而不是盲目的 Ctrl+C，Ctrl+V，切记切记！！

实验五

实验五主要考察的内容如下：
对于聚集函数 sum, max, count 的使用，同时有无 group by 的意识；

对于分部分查询的熟练程度；（可能会有其他方法，但这部分我分块查询用的比较多~~）

对于 union all 的了解及区分 union all 和 union 的区别；

create table test5_01 as
	select distinct substr(name, 2, length(name)) first_name, count(*) frequency
	from pub.student
	group by substr(name, 2, length(name))

create table test5_02
	(letter varchar(2),
	frequency numeric(4))

insert into test5_02
select letter, count(*) frequency
from ((select substr(name, 2, 1) letter
	   from pub.student
	   where substr(name, 2, 1) is not null)
	 union all
	  (select substr(name, 3, 1) letter
	   from pub.student
	   where substr(name, 3, 1) is not null))
group by letter

create table test5_03
	(dname varchar(30),
	class varchar(10),
	p_count1 int,
	p_count2 int,
	p_count int)

-----------一张表一张表地看思路更清晰哦（一共就 t1，t2 两张表）--------------
insert into test5_03
select t1.dname, t1.class, t2.p_count1, (t1.p_count - t2.p_count1) p_count2, t1.p_count
from 
	(select distinct dname, class, count(*) p_count
	from pub.student
	where dname is not null
	group by dname, class) t1,					--t1 表：找到 dname, class, p_count
	
	(select distinct dname, class, count(*) p_count1
	from 
		(select sid, dname, class, sum(credit) sum_credit	--找到总学分
		from 
			(select sid, cid, dname, class, max(score) max_score	--找到成绩的最大值
			from pub.student_course natural join pub.student
			group by sid, cid, dname, class) natural join pub.course
		where max_score >= 60					--最大成绩 >= 60 才计入学分
		and dname is not null
		group by sid, dname, class)
	where sum_credit >= 10						--总学分 > 10 才选出来
	group by dname, class) t2,					--t2 表：找到 p_count1（找 dname 和 class 是为了在后面进行连接）

where t1.dname = t2.dname
and t1.class = t2.class

现在给出计算 p_count2 的代码，有兴趣可以看看哈~~

select distinct dname, class, count(*) p_count2
from 
	((select sid, dname, class, sum(credit) sum_credit		--选了课但是没有达标的学生
	from 
		(select sid, cid, dname, class, max(score) max_score
		from pub.student_course natural join pub.student
		group by sid, cid, dname, class) natural join pub.course
	where max_score >= 60
	and dname is not null
	group by sid, dname, class)
union
	(select distinct sid, dname, class, 0 as sum_credit		--在 pub.student 中但是不在 pub.student_course 中的学生，这部分学生也算作不达标哦~~
	from pub.student
	where sid not in
		(select sid
		from pub.student_course)))
where sum_credit < 10
group by dname, class

create table test5_04
	(dname varchar(30),
	class varchar(10),
	p_count1 int,
	p_count2 int,
	p_count int)

-----------一张表一张表地看思路更清晰哦（一共就 t1，t2 两张表）--------------
insert into test5_04
select t1.dname, t1.class, t2.p_count1, (t1.p_count - t2.p_count1) p_count2, t1.p_count
from 
	(select distinct dname, class, count(*) p_count
	from pub.student
	where dname is not null
	group by dname, class) t1,
	
	(select distinct dname, class, count(*) p_count1
	from 
		(select sid, dname, class, sum(credit) sum_credit
		from 
			(select sid, cid, dname, class, max(score) max_score
			from pub.student_course natural join pub.student
			group by sid, cid, dname, class) natural join pub.course
		where max_score >= 60
		and dname is not null
		group by sid, dname, class)
	where sum_credit >= 					--Look at here! 增加的条件在这里 —— 就是根据 class 来决定 sum_credit 的判定条件哦
		case
			when class <= 2008 then 8
			when class > 2008 then 10
		end
	group by dname, class) t2
	
where t1.dname = t2.dname
and t1.class = t2.class

手动算 p_count2 也是在相同的地方加上 case 即可。

create table test5_05
	(dname varchar(20),
	avg_ds_score int,
	avg_os_score int);

insert into test5_05
select distinct t1.dname, t2.avg_ds_score, t3.avg_os_score
from
	(select distinct dname
	from pub.student) t1,
	(select distinct dname, round(avg_ds_score, 0) avg_ds_score
	from 
		(select distinct dname, avg(max_ds_score) avg_ds_score
		from 
			(select distinct sid, dname, max(score) max_ds_score
			from pub.student natural join pub.student_course
			where cid = (select cid from pub.course where name = '数据结构')
			group by sid, dname)
		group by dname)) t2,
	(select distinct dname, round(avg_os_score, 0) avg_os_score
	from 
		(select distinct dname, avg(max_os_score) avg_os_score
		from 
			(select distinct sid, dname, max(score) max_os_score
			from pub.student natural join pub.student_course
			where cid = (select cid from pub.course where name = '操作系统')
			group by sid, dname)
		group by dname)) t3
where t1.dname = t2.dname
and t2.dname = t3.dname

create table test5_06 as
	select t1.sid, t1.name, '计算机科学与技术学院' as dname, t2.ds_score, t3.os_score
	from 
		(select sid, name
		from pub.student S
		where dname = '计算机科学与技术学院'
		and not exists
		(	(select cid
			from pub.course
			where name = '数据结构' or name = '操作系统')
		minus
			(select cid
			from pub.student_course T
			where S.sid = T.sid))) t1,
			
		(select sid, max(score) ds_score
		from pub.student_course
		where cid = (select cid from pub.course where name = '数据结构')
		group by sid) t2,
		
		(select sid, max(score) os_score
		from pub.student_course
		where cid = (select cid from pub.course where name = '操作系统')
		group by sid) t3
		
	where t1.sid = t2.sid
	and t2.sid = t3.sid

create table test5_07 as
	select sid, name, dname, ds_score, os_score
	from 
		(select distinct sid, name, dname
		from pub.student natural join pub.student_course
		where dname = '计算机科学与技术学院'
		and cid in (select cid from pub.course where name = '数据结构' or name = '操作系统')) 
	natural left outer join 
		(select distinct sid, max(score) ds_score
		from pub.student_course natural join pub.course
		where name = '数据结构'
		group by sid)
	natural left outer join
		(select distinct sid, max(score) os_score
		from pub.student_course natural join pub.course
		where name = '操作系统'
		group by sid)

create table test5_08 as
	select sid, name, dname, ds_score, os_score
	from 
		(select distinct sid, name, dname
		from pub.student
		where dname = '计算机科学与技术学院') 
	natural left outer join 
		(select distinct sid, max(score) ds_score
		from pub.student_course natural join pub.course
		where name = '数据结构'
		group by sid)
	natural left outer join
		(select distinct sid, max(score) os_score
		from pub.student_course natural join pub.course
		where name = '操作系统'
		group by sid)

再次强调：一定是看懂思路之后自己实践哈~~
有问题还请斧正！
在这里插入图片描述

标签：dname,pub,score,2020,sid,数据库系统,山东大学,class,select
来源： https://blog.csdn.net/qq_53413759/article/details/120538349