山东大学软件学院2017-18学年数据库实验五、六、七、八

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，答案不唯一可能有些效率不高，仅供参考。

• 实验五

实验步骤：（不要求罗列完整源代码） create table test5_00 as select * from pub.TEACHER grant all on test5_00 to user201600301315 update test5_00 set age = 88                          select * from test5_00                                  Commit Rollback update test5_00 set age = age + 1 Rollback Commit update test5_00 set age = age + 2 Commit select * from test5_00                                  Rollback select * from userb201600301315.test5_00               update test5_00 set age = age - 2 update test5_00 set age = age - 2 select * from test5_00                                  select * from userb201600301315.test5_00              Commit select * from userb201600301315.test5_00               Rollback update userb201600301315.test5_00 set age = age – 10 Select * from test5_00                                  create table test5_01 as select * from test5_00 Rollback select * from userb201600301315.test5_00             select * from userb201600301315.test5_00              Rollback select * from userb201600301315.test5_00               结果为：88 90 90 86 90 90 86 86 76 86

• 实验六

实验步骤：（不要求罗列完整源代码） 找出年龄小于20岁且是“物理学院”的学生的学号、姓名、院系名称，按学号排序   create view test6_01 as select sid,name,dname from pub.STUDENT where age<20 and dname='物理学院' order by sid   查询统计2009级、软件学院的每个学生的学号、姓名、总成绩（sum_score）   create view test6_02 as select pub.STUDENT.SID sid,pub.STUDENT.NAME name,sum(score) sum_score from pub.STUDENT,pub.STUDENT_COURSE where pub.STUDENT.SID=pub.STUDENT_COURSE.SID and pub.STUDENT.CLASS=2009 and pub.STUDENT.DNAME='软件学院' group by pub.STUDENT.SID,pub.STUDENT.NAME   查询统计2010级、计算机科学与技术学院、操作系统的学生成绩表，内容有学号姓名成绩。   create view test6_03 as select pub.STUDENT.SID sid,pub.STUDENT.NAME name,pub.STUDENT_COURSE.SCORE score from pub.STUDENT,pub.STUDENT_COURSE,pub.COURSE where pub.STUDENT_COURSE.CID=pub.COURSE.CID and pub.STUDENT.SID=pub.STUDENT_COURSE.SID and pub.STUDENT.CLASS=2010 and pub.STUDENT.DNAME='计算机科学与技术学院' and pub.COURSE.NAME = '操作系统'   找出选修“数据库系统”课程且成绩大于90学生的学号、姓名   create view test6_04 as select pub.STUDENT.SID sid,pub.STUDENT.NAME name from pub.STUDENT,pub.STUDENT_COURSE,pub.COURSE where pub.STUDENT_COURSE.CID=pub.COURSE.CID and pub.STUDENT.SID=pub.STUDENT_COURSE.SID and pub.COURSE.NAME = '数据库系统' and pub.STUDENT_COURSE.SCORE>90   找出姓名叫“李龙”的学生的学号及其选修全部课程的课程号、课程名和成绩   create view test6_05 as select pub.STUDENT.SID sid,pub.STUDENT_COURSE.CID cid,pub.COURSE.NAME,pub.STUDENT_COURSE.SCORE from pub.STUDENT,pub.STUDENT_COURSE,pub.COURSE where pub.STUDENT_COURSE.CID=pub.COURSE.CID and pub.STUDENT.SID=pub.STUDENT_COURSE.SID and pub.STUDENT.NAME='李龙'   找出选修了所有课程的学生的学号、姓名   create or replace view test6_06 as select a.SID,a.NAME from pub.STUDENT a where not exists ((select cid from pub.COURSE) minus (select cid from  pub.STUDENT natural join pub.STUDENT_COURSE where a.sid = sid)）   找出选修了所有课程并且所有课程全部通过的学生的姓名、学号   create or replace view test6_07 as select a.SID,a.NAME from pub.STUDENT a where not exists( (select cid from pub.COURSE) minus (select cid from  pub.STUDENT natural join pub.STUDENT_COURSE where a.sid = sid and score>=60)）   检索先行课的学分为2的课程号、课程名   create or replace view test6_08 as select A.cid, A.NAME from pub.COURSE A, pub.COURSE B where A.FCID = B.CID and B.CREDIT = 2   查询统计2010级、化学与化工学院的学生总学分表，内容有学号、姓名、总学分sum_credit   注：学分为成绩大于等于60才算   create or replace view test6_09 as select pub.STUDENT.SID sid,pub.STUDENT.NAME name,sum(credit) sum_credit 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.STUDENT.CLASS=2010 and pub.STUDENT.DNAME='化学与化工学院' and pub.STUDENT_COURSE.SCORE>=60                      group by pub.STUDENT.SID,pub.STUDENT.NAME   找出有间接先行课的所有课程的课程号、课程名称   create or replace view test6_10 as select A.CID,A.NAME from pub.COURSE A,pub.COURSE B,pub.COURSE C where A.FCID = B.CID and B.FCID = C.CID

• 实验七

实验步骤：（不要求罗列完整源代码） 统计名字的使用频率   create table test7_01 as select First_name, (count(*)) frequency from (select (substr(name,2)) First_name from pub.STUDENT) group by First_name   2.统计名字每个字的使用频率   create table test7_02 as select letter,(count(*)) frequency from ((select (substr(name,2,1)) letter from pub.STUDENT) union all (select (substr(name,3,1)) letter from pub.STUDENT where substr(name,3,1) is not NULL)) group by letter   3.统计学院班级学分达标情况统计表1   create table test7_03 as select * from (select dname, class, (count(*)) P_count1  from (select S.sid, S.DNAME, S.CLASS, sum_credit     from pub.STUDENT S,             (select sid, (sum(credit)) sum_credit          from pub.STUDENT_COURSE SC, pub.COURSE C          where SC.CID = C.CID          and SC.SCORE >= 60          and dname is not NULL          group by sid) B     where S.SID = B.SID)  where sum_credit > = 10  and dname is not NULL  group by dname, class)   natural full outer join   (select dname, class, (count(*)) P_count2  from (select S.sid, S.DNAME, S.CLASS, sum_credit        from pub.STUDENT S,             (select sid, (sum(credit)) sum_credit          from pub.STUDENT_COURSE SC, pub.COURSE C          where SC.CID = C.CID          and SC.SCORE >= 60              and dname is not NULL          group by sid) B        where S.SID = B.SID)  where sum_credit < 10  and dname is not NULL  group by dname, class)   natural full outer join   (select *  from (select dname, class, (count(*)) P_count        from pub.STUDENT        where dname is not NULL        group by dname, class) sum_count)   4.统计学院班级学分达标情况统计表2   create table test7_04 as select * from (select dname, class, (count(*)) P_count1     from (select A.sid, A.DNAME, A.CLASS, sum_credit        from pub.STUDENT A, (select sid, (sum(credit)) sum_credit        from pub.STUDENT_COURSE S, pub.COURSE K        where S.CID = K.CID        and S.SCORE >= 60        group by sid) B        where A.SID = B.SID and A.DNAME is not NULL)        where (sum_credit > = 10)     or (sum_credit > = 8 and class <= 2008)   group by dname, class) natural full outer join (select dname, class, (count(*)) P_count2     from (select A.sid, A.DNAME, A.CLASS, sum_credit        from pub.STUDENT A, (select sid, (sum(credit)) sum_credit        from pub.STUDENT_COURSE S, pub.COURSE K        where S.CID = K.CID        and S.SCORE >= 60        group by sid) B        where A.SID = B.SID)     where sum_credit < 10     and dname is not NULL     group by dname, class) natural full outer join (select * from (select dname, class, (count(*)) P_count     from pub.STUDENT     where dname is not NULL     group by dname, class) sum_count)

• 实验八

实验步骤：（不要求罗列完整源代码） 查询各院系的数据结构平均成绩和操作系统平均成绩。 create table test8_01 as select * from (select dname, (round(avg(s1),0)) Avg_ds_score from (select pub.STUDENT.SID, max(score) s1 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' group by pub.STUDENT.SID) A, pub.STUDENT B where A.SID = B.SID and dname is not NULL group by dname)   natural full outer join   (select dname, (round(avg(s2),0)) Avg_os_score   from (select pub.STUDENT.SID, max(score) s2          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'          group by pub.STUDENT.SID) A,          pub.STUDENT B   where A.SID = B.SID   and dname is not NULL   group by dname) 查询计算机科学与技术学院同时选修了数据结构、操作系统两门课的学生详细信息 create table test8_02 as select A.SID, name, dname, ds_score, os_score from (select pub.STUDENT.SID, max(score) ds_score 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' group by pub.STUDENT.SID) A,   (select pub.STUDENT.SID, max(score) os_score 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' group by pub.STUDENT.SID) B,   pub.STUDENT C   where A.SID = B.SID and B.SID = C.SID and dname = '计算机科学与技术学院' 查询计算机科学与技术学院选修了数据结构或操作系统的学生详细信息 create table test8_03 as select sid, name, dname, ds_score, os_score from (select * from (select pub.STUDENT.SID, max(score) ds_score 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' group by pub.STUDENT.SID)   natural full outer join   (select pub.STUDENT.SID, max(score) os_score 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' group by pub.STUDENT.SID) )    natural join pub.STUDENT where dname = '计算机科学与技术学院'   查询计算机科学与技术学院所有学生的信息   create table test8_04 as select sid, name, dname, ds_score, os_score from (select * from (select pub.STUDENT.SID, max(score) ds_score 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' group by pub.STUDENT.SID)   natural full outer join   (select pub.STUDENT.SID, max(score) os_score 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' group by pub.STUDENT.SID) )  natural right outer join pub.STUDENT where dname = '计算机科学与技术学院'