山东大学软件学院2017-18学年数据库实验一、二、三、四
注：所有实验均为自己思考所答，非转载抄袭，并全部通过实验系统检测。希望学弟学妹自主思考，答案不唯一，仅供参考。
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实验一
创建学生信息表test1_student（学生编号、姓名、性别、年龄、出生日期、院系名称、班级）
create table test1_student
( sid char(12) not null,
name varchar2(10) not null,
sex char(2),
age int,
birthday date,
dname varchar2(30),
class varchar2(10) )
创建教师信息表test1_teacher（教师编号、姓名、性别、年龄、院系名称）
create table test1_teacher
( tid char(6) not null,
name varchar2(10) not null,
sex char(2),
age int,
dname varchar2(30))
创建课程信息表test1_course（课程编号、课程名称、先行课编号、学分）
create table test1_course
( cid char(6) not null,
name varchar2(40) not null,
fcid char(6),
credit numeric (4,1))
创建学生选课信息表test1_student_course（学号、课程号、成绩、教师编号）
create table test1_student_course
( sid char(12) not null,
cid char(6) not null,
score numeric (5,1),
tid char (6))
创建教师授课信息表test1_teacher_course（教师编号、课程编号）
create table test1_teacher_course
( tid char(6) not null,
cid char(6) not null)
给表test1_student插入学生信息
insert into test1_student values
(200800020101,'王欣','女',19,date'1994-02-02','计算机学院','2010')
insert into test1_student values
(200800020102,'李华','女',20,date'1995-03-03','软件学院','2009')
insert into test1_student values
(200800020103,'赵岩','男',21,date'1996-04-04','软件学院','2009')
给表test1_teacher插入教师信息
insert into test1_teacher values(100101,'张老师','男',44,'计算机学院')
insert into test1_teacher values(100102,'李老师','女',45,'软件学院')
insert into test1_teacher values(100103,'马老师','男',46,'计算机学院')
给表test1_course插入三行数据
insert into test1_course values(300001,'数据结构',null,2)
insert into test1_course values(300002,'数据库',300001,2.5)
insert into test1_course values(300003,'操作系统',300001,4)
给表test1_student_course插入三行数据
insert into test1_student_course values(200800020101,300001,91.5,100101)
insert into test1_student_course values(200800020101,300002,92.6,100102)
insert into test1_student_course values(200800020101,300003,93.7,100103)
给表test1_teacher_course插入三行数据
insert into test1_teacher_course values(100101,300001)
insert into test1_teacher_course values(100102,300002)
insert into test1_teacher_course values(100103,300003)
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实验二
找出没有选修任何课的学生的学号、姓名（即没有选课记录的学生）
采用集合差运算
create table test2_01 as
(select sid,name from pub.STUDENT)
minus
( select pub.STUDENT.sid,pub.STUDENT.name
from pub.STUDENT,pub.STUDENT_COURSE
where (pub.STUDENT.SID)=(pub.STUDENT_COURSE.SID))
找出至少选修了学号为”200900130417”的学生所选修的一门课的学生的学号、姓名
采用多个in子查询
create table test2_02 as
( select sid,name from pub.STUDENT
where (pub.STUDENT.SID)
in ( select sid
from pub.STUDENT_COURSE
where cid
in ( select cid
from pub.STUDENT_COURSE
where sid=200900130417 ) ) )
找出至少选修了一门其先行课程号为“300002”号课程的学生的学号、姓名。
同2题采用多个in子查询
create table test2_03 as
( select sid,name
from pub.STUDENT
where sid
in ( select sid
from pub.STUDENT_COURSE
where cid
in ( select cid
from pub.COURSE
where fcid=300002 ) ) )
找出选修了“操作系统”并且也选修了“数据结构”的学生的学号、姓名。
集合的并集运算
create table test2_04 as
((select pub.STUDENT.SID,pub.STUDENT.NAME
from pub.STUDENT,pub.STUDENT_COURSE,pub.COURSE
where (pub.STUDENT.SID)=(pub.STUDENT_COURSE.SID)
and (pub.STUDENT_COURSE.CID)=(pub.COURSE.CID)
and pub.COURSE.NAME='操作系统')
intersect
( select pub.STUDENT.SID,pub.STUDENT.NAME
from pub.STUDENT,pub.STUDENT_COURSE,pub.COURSE
where (pub.STUDENT.SID)=(pub.STUDENT_COURSE.SID)
and (pub.STUDENT_COURSE.CID)=(pub.COURSE.CID)
and pub.COURSE.NAME='数据结构'))
查询20岁的所有有选课的学生的学号、姓名、平均成绩（avg_score，此为列名，下同）（平均成绩四舍五入到个位）、总成绩（sum_score）
Round（）函数可以四舍五入
create table test2_05 as
( select pub.STUDENT.SID sid,pub.STUDENT.NAME name,round(avg(pub.STUDENT_COURSE.SCORE)) avg_score,sum(pub.STUDENT_COURSE.SCORE) sum_score
from pub.STUDENT,pub.STUDENT_COURSE
where pub.STUDENT.AGE='20'
and (pub.STUDENT.SID)= (pub.STUDENT_COURSE.SID)
group by pub.STUDENT.SID,pub.STUDENT.NAME)
查询所有课以及这门课的最高成绩，test2_06有两个列：课程号cid、最高成绩max_score
create table test2_06 as
( select cid,max(pub.STUDENT_COURSE.SCORE) max_score
from pub.STUDENT_COURSE
group by cid)
查询所有不姓张、不姓李、也不姓王的学生的学号sid、姓名name
运用like查询张李王开头的名字
create table test2_07 as
( ( select sid,name
from pub.STUDENT)
minus
( select sid,name
from pub.STUDENT
where name like '张%'
or name like'李%'
or name like'王%'))
查询学生表中每一个姓氏及其人数（不考虑复姓），text2_08有两个列：second_name、p_count
运用substr（）函数来截取姓氏
create table test2_08 as
( select (substr(name, 0, 1)) second_name, (count(*)) p_count
from pub.STUDENT
group by substr(name, 0, 1))
查询选修了300003号课程的学生的sid、name、score
create table test2_09 as
(select pub.STUDENT.SID sid,pub.STUDENT.NAME name,pub.STUDENT_COURSE.SCORE score
from pub.STUDENT,pub.STUDENT_COURSE
where (pub.STUDENT.SID)=(pub.STUDENT_COURSE.SID)
and (pub.STUDENT_COURSE.CID)=300003)
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实验三
将pub用户下的Student_31及数据复制到主用户的表test3_01,删除表中的学号不全是数字的那些错误数据，学号应该是数字组成，不能够包含字母空格等非数字字符。
方法一：（不用delete，运用正则表达式直接选出全是数字的数据）
create table test3_01 as
( select * from pub.STUDENT_31
where regexp_like( sid,'^[[:digit:]]{12}$'))
方法二：（运用delete删除那些不全是数字的数据）
create table test3_01 as( select * from pub.STUDENT_31 )
delete from test3_01
where sid in
( select sid from pub.STUDENT_31 where sid not in
( select sid from pub.STUDENT_31 where regexp_like(sid,'^[[:digit:]]{12}$')))
将pub用户下的Student_31及数据复制到主用户的表test3_02,删除表中的出生日期和年龄（截止到2012年的年龄，即年龄=2012-出生年份）不一致的那些错误数据。
运用extract（）函数将birthday中的年份提取
运用decode（）函数将age与2012-extract（）进行比较，相同为1不同为0，并更名为compare，删除为0的那些数据
create table test3_02 as(select * from pub.STUDENT_31)
delete from test3_02 where sid in
( select sid from test3_02,( select sid a_sid,decode(age,2012-extract(year from birthday),1,0) compare from test3_02) where sid = a_sid and compare = 0)
将pub用户下的Student_31及数据复制到主用户的表test3_03,删除表中的性别有错误的那些错误数据（性别只能够是“男”、“女”或者空值）。
create table test3_03 as(select * from pub.STUDENT_31)
delete from test3_03 where sex not in
(select sex from test3_03 where sex = '男' or sex='女' or sex is null)
将pub用户下的Student_31及数据复制到主用户的表test3_04,删除表中的院系名称有空格的、院系名称为空值的或者院系名称小于3个字的那些错误数据。
运用length（）函数判断字符串长短
运用trim（）函数去掉字符串前后的空格
运用replace（）函数去掉字符串中间的空格
create table test3_04 as(select * from pub.STUDENT_31)
delete from test3_04 where sid in
( select sid from test3_04
where dname is null
or length(dname)<3
or length(dname) > length(trim(dname))
or length(dname) > length(replace(dname,' ','') ) )
将pub用户下的Student_31及数据复制到主用户的表test3_05,删除表中的班级不规范的那些错误数据，不规范是指和大多数不一致。
通过观察表的内容解决实际问题，发现大部分为XXXX四位数，所以删除长度大于四的数据即可
create table test3_05 as(select * from pub.STUDENT_31)
delete from test3_05 where length(class) > 4
将pub用户下的Student_31及数据复制到主用户的表test3_06,删除表中的错误数据，不规范的数据也被认为是那些错误数据。
将之前5个实验题中的语句依次执行即可
create table test3_06 as( select * from pub.STUDENT_31 )
delete from test3_06 where sid in (select sid from pub.STUDENT_31 where sid not in (select sid from pub.STUDENT_31 where regexp_like(sid,'^[[:digit:]]{12}$')))
delete from test3_06 where sid in (select sid from test3_06,(select sid a_sid,decode(age,2012-extract(year from birthday),1,0) compare from test3_06) where sid = a_sid and compare = 0)
delete from test3_06 where name in(select name from test3_06 where length(name)<2 or length(name)>length(trim(name)) or length(name) > length(replace(name,' ',''))）
delete from test3_06 where sex <>'男' and sex<>'女' and sex is not null
delete from test3_06 where sid in(select sid from test3_06 where dname is null or length(dname)<3 or length(dname)>length(trim(dname)) or length(dname) > length(replace(dname,' ','')) )
delete from test3_06 where length(class)>4
将pub用户下的Student_32及数据复制到主用户的表test3_07,删除其中的错误数据，错误指如下情况：学号在学生信息pub.student中不存在的；
create table test3_07 as( select * from pub.STUDENT_COURSE_32)
delete from test3_07 where sid not in ( select sid from pub.student )
将pub用户下的Student_32及数据复制到主用户的表test3_08,删除其中的错误数据，错误指如下情况：课程号和教师编号在教师授课表pub.teacher_course中不同时存在的，即没有该教师教该课程；
Where后可以用括号来多项同时查询
create table test3_08 as( select * from pub.STUDENT_COURSE_32 )
delete from test3_08 where (tid,cid) not in ( select tid,cid from pub.TEACHER_COURSE )
将pub用户下的Student_32及数据复制到主用户的表test3_09,删除其中的错误数据，错误数据指如下情况：成绩数据有错误（需要先找到成绩里面的错误）。
成绩正确数据是0-100分之内的，所以用between语句
create table test3_09 as( select * from pub.STUDENT_COURSE_32 )
delete from test3_09 where score not in (select score from test3_09 where score between 0 and 100)
将pub用户下的Student_32及数据复制到主用户的表test3_10,删除其中的错误数据，错误指如下情况：
将上题语句依次执行
create table test3_10 as( select * from pub.STUDENT_COURSE_32 )
delete from test3_10 where sid not in ( select sid from pub.student )
delete from test3_10 where cid not in ( select cid from pub.COURSE )
delete from test3_10 where tid not in ( select tid from pub.TEACHER )
delete from test3_10 where (tid,cid) not in ( select tid,cid from pub.TEACHER_COURSE )
delete from test3_10 where score not in (select score from test3_10 where score between 0 and 100)
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实验四
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create table test4_01 as select * from pub.STUDENT_41 alter table test4_01 add (sum_score numeric(6, 1)) update test4_01 set sum_score = (select sum(score) from pub.STUDENT_COURSE where pub.STUDENT_COURSE.SID = test4_01.SID);
create table test4_02 as select * from pub.STUDENT_41 alter table test4_02 add (avg_score numeric(4, 1)) update test4_02 set avg_score = (select round(avg(score),1) from pub.STUDENT_COURSE where pub.STUDENT_COURSE.SID = test4_02.SID);
create table test4_03 as select * from pub.STUDENT_41 alter table test4_03 add (sum_credit numeric(4, 1)) update test4_03 set sum_credit = (select sum(credit) from pub.COURSE, pub.STUDENT_COURSE where pub.STUDENT_COURSE.SID = test4_03.SID and pub.STUDENT_COURSE.SCORE >= 60 and pub.STUDENT_COURSE.CID = pub.COURSE.CID);
create table test4_04 as select * from pub.STUDENT_41 update test4_04 set dname = (select did from pub.DEPARTMENT where test4_04.DNAME = pub.DEPARTMENT.DNAME) where exists(select * from pub.DEPARTMENT where test4_04.DNAME = pub.DEPARTMENT.DNAME)
create table test4_05 as select * from pub.STUDENT_41 alter table test4_05 add (sum_score numeric(6, 1)) update test4_05 set sum_score = (select sum(score) from pub.STUDENT_COURSE where pub.STUDENT_COURSE.SID = test4_05.SID);
alter table test4_05 add (avg_score numeric(4, 1)) update test4_05 set avg_score = (select round(avg(score),1) from pub.STUDENT_COURSE where pub.STUDENT_COURSE.SID = test4_05.SID);
alter table test4_05 add (sum_credit numeric(4, 1)) update test4_05 set sum_credit = (select sum(credit) from pub.COURSE, pub.STUDENT_COURSE where pub.STUDENT_COURSE.SID = test4_05.SID and pub.STUDENT_COURSE.SCORE >= 60 and pub.STUDENT_COURSE.CID = pub.COURSE.CID);
alter table test4_05 add (did varchar(2)) update test4_05 set did= (select did from pub.DEPARTMENT where test4_05.DNAME = pub.DEPARTMENT.DNAME) where exists(select * from pub.DEPARTMENT where test4_05.DNAME = pub.DEPARTMENT.DNAME)
update test4_05 set did = (select did from pub.DEPARTMENT_41 where test4_05.DNAME = pub.DEPARTMENT_41.DNAME) where exists(select * from pub.DEPARTMENT_41 where test4_05.DNAME = pub.DEPARTMENT_41.DNAME)
update test4_05 set did = '00' where did is null
create table test4_06 as select * from pub.STUDENT_42 update test4_06 set name = (replace(name, ' ', ''))
create table test4_07 as select * from pub.STUDENT_42 update test4_07 set sex = (replace(sex, ' ', '')) update test4_07 set sex = (replace(sex, '性', ''))
create table test4_08 as select * from pub.STUDENT_42 update test4_08 set class = (replace(class, '级', ''))
create table test4_09 as select * from pub.STUDENT_42 update test4_09 set age = (2012 - extract(year from birthday)) where age is NULL
create table test4_10 as select * from pub.STUDENT_42 update test4_10 set name = (replace(name, ' ', '')) update test4_10 set dname = (replace(dname, ' ', '')) update test4_10 set sex = (replace(sex, ' ', '')) update test4_10 set sex = (replace(sex, '性', '')) update test4_10 set class = (replace(class, '级', '')) update test4_10 set age = (2012 - extract(year from birthday)) where age is NULL
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