• oracle sql试题


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    数据准备

    create table student(
      sno varchar2(10) primary key,
      sname varchar2(20),
      sage number(3),
      ssex varchar2(5)
    );


    create table teacher(
     tno varchar2(10) primary key,
     tname varchar2(20)
    );


    create table course(
      cno varchar2(10),
      cname varchar2(20),
      tno varchar2(20),
      constraint pk_course primary key (cno,tno)
    );


    create table sc(
       sno varchar2(10),
       cno varchar2(10),
       score number(4,2),
       constraint pk_sc primary key (sno,cno)
    );

    /*******初始化学生表的数据******/
    insert into student values ('s001','张三',23,'男');
    insert into student values ('s002','李四',23,'男');
    insert into student values ('s003','吴鹏',25,'男');
    insert into student values ('s004','琴沁',20,'女');
    insert into student values ('s005','王丽',20,'女');
    insert into student values ('s006','李波',21,'男');
    insert into student values ('s007','刘玉',21,'男');
    insert into student values ('s008','萧蓉',21,'女');
    insert into student values ('s009','陈萧晓',23,'女');
    insert into student values ('s010','陈美',22,'女');
    commit;


    /******************初始化教师表***********************/
    insert into teacher values ('t001', '刘阳');
    insert into teacher values ('t002', '谌燕');
    insert into teacher values ('t003', '胡明星');
    commit;


    /***************初始化课程表****************************/
    insert into course values ('c001','J2SE','t002');
    insert into course values ('c002','Java Web','t002');
    insert into course values ('c003','SSH','t001');
    insert into course values ('c004','Oracle','t001');
    insert into course values ('c005','SQL SERVER 2005','t003');
    insert into course values ('c006','C#','t003');
    insert into course values ('c007','JavaScript','t002');
    insert into course values ('c008','DIV+CSS','t001');
    insert into course values ('c009','PHP','t003');
    insert into course values ('c010','EJB3.0','t002');
    commit;
    /***************初始化成绩表***********************/
    insert into sc values ('s001','c001',78.9);
    insert into sc values ('s002','c001',80.9);
    insert into sc values ('s003','c001',81.9);
    insert into sc values ('s004','c001',60.9);
    insert into sc values ('s001','c002',82.9);
    insert into sc values ('s002','c002',72.9);
    insert into sc values ('s003','c002',81.9);
    insert into sc values ('s001','c003','59');
    commit; 


    student table

    sno

    sname

    sage

    ssex

    s001

    张三

    23

    s002

    李四

    23

    s003

    吴鹏

    25

    s004

    琴沁

    20

    s005

    王丽

    20

    s006

    李波

    21

    s007

    刘玉

    21

    s008

    萧蓉

    21

    s009

    陈萧晓

    23

    s010

    陈美

    22

    teacher table

    tno

    tname

    t001

    刘阳

    t002

    胡燕

    t003

    胡明星

    course table

    cno

    cname

    tno

    c001

    J2SE

    t002

    c002

    Java Web

    t002

    c003

    SSH

    t001

    c004

    Oracle

    t001

    c005

    SQL SERVER 2005

    t003

    c006

    C#

    t003

    c007

    JavaScript

    t002

    c008

    DIV+CSS

    t001

    c009

    PHP

    t003

    c010

    EJB3.0

    t002

    score table

    sno

    cno

    score

    s001

    c001

    78.9

    s002

    c001

    80.9

    s003

    c001

    81.9

    s004

    c001

    60.9

    s001

    c002

    82.9

    s002

    c002

    72.9

    s003

    c002

    81.9

    s001

    c003

    59

     试题

    1、查询“c001”课程比“c002”课程成绩高的所有学生的学号;

    2、查询平均成绩大于60分的同学的学号和平均成绩;

    3、查询所有同学的学号、姓名、选课数、总成绩;

    4、查询姓“刘”的老师的个数;

    5、查询没学过“谌燕”老师课的同学的学号、姓名;

    6、查询学过“c001”并且也学过编号“c002”课程的同学的学号、姓名;

    7、查询学过“谌燕”老师所教的所有课的同学的学号、姓名;

    8、查询课程编号“c002”的成绩比课程编号“c001”课程低的所有同学的学号、姓名;

    9、查询所有课程成绩小于60 分的同学的学号、姓名;

    10、查询没有学全所有课的同学的学号、姓名;

    11、查询至少有一门课与学号为“s001”的同学所学相同的同学的学号和姓名;

    12、查询至少学过学号为“s001”同学所有一门课的其他同学学号和姓名;
    13、把“SC”表中“谌燕”老师教的课的成绩都更改为此课程的平均成绩;
    14、查询和“s001”号的同学学习的课程完全相同的其他同学学号和姓名;
    15、删除学习“谌燕”老师课的SC 表记录;
    16、向SC 表中插入一些记录,这些记录要求符合以下条件:没有上过编号“c002”课程的同学学号、“c002”号课的平均成绩;
    17、查询各科成绩最高和最低的分:以如下形式显示:课程ID,最高分,最低分
    18、按各科平均成绩从低到高和及格率的百分数从高到低顺序
    19、查询不同老师所教不同课程平均分从高到低显示
    20、统计列印各科成绩,各分数段人数:课程ID,课程名称,[100-85],[85-70],[70-60],[ <60]
    21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
    22、查询每门课程被选修的学生数
    23、查询出只选修了一门课程的全部学生的学号和姓名
    24、查询男生、女生人数
    25、查询姓“张”的学生名单
    26、查询同名同性学生名单,并统计同名人数
    27、1981 年出生的学生名单(注:Student 表中Sage 列的类型是number)
    28、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列
    29、查询平均成绩大于85 的所有学生的学号、姓名和平均成绩
    30、查询课程名称为“数据库”,且分数低于60 的学生姓名和分数
    31、查询所有学生的选课情况;
    32、查询任何一门课程成绩在70 分以上的姓名、课程名称和分数;
    33、查询不及格的课程,并按课程号从大到小排列
    34、查询课程编号为c001 且课程成绩在80 分以上的学生的学号和姓名;
    35、求选了课程的学生人数
    36、查询选修“谌燕”老师所授课程的学生中,成绩最高的学生姓名及其成绩
    37、查询各个课程及相应的选修人数
    38、查询不同课程成绩相同的学生的学号、课程号、学生成绩
    39、查询每门功课成绩最好的前两名
    40、统计每门课程的学生选修人数(超过10 人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,若人数相同,按课程号升序排列
    41、检索至少选修两门课程的学生学号
    42、查询全部学生都选修的课程的课程号和课程名
    43、查询没学过“谌燕”老师讲授的任一门课程的学生姓名
    44、查询两门以上不及格课程的同学的学号及其平均成绩
    45、检索“c004”课程分数小于60,按分数降序排列的同学学号
    46、删除“s002”同学的“c001”课程的成绩

    答案

     
    1.
    *********************************
    select a.* from
    (select * from sc a where a.cno='c001') a,
    (select * from sc b where b.cno='c002') b
    where a.sno=b.sno and a.score > b.score;
    *********************************
    select * from sc a
    where a.cno='c001'
    and  exists(select * from sc b where b.cno='c002' and a.score>b.score
    and a.sno = b.sno)
    *********************************
    2.
    *********************************
    select sno,avg(score) from sc  group by sno having avg(score)>60;
    *********************************
    3.
    *********************************
    select a.*,s.sname from (select sno,sum(score),count(cno) from sc group by sno) a ,student s where a.sno=s.sno
    *********************************
    4.
    *********************************
    select count(*) from teacher where tname like '刘%';
    *********************************
    5.
    *********************************
    select a.sno,a.sname from student a
    where a.sno
    not in
    (select distinct s.sno
     from sc s,
          (select c.*
           from course c ,
               (select tno
                from teacher t
                where tname='谌燕')t
           where c.tno=t.tno) b
      where s.cno = b.cno )
    *********************************
    select * from student st where st.sno not in
    (select distinct sno from sc s join course c on s.cno=c.cno
    join teacher t on c.tno=t.tno where tname='谌燕')
    *********************************
    6.
    *********************************
    select st.* from sc a
    join sc b on a.sno=b.sno
    join student st
    on st.sno=a.sno
    where a.cno='c001' and b.cno='c002' and st.sno=a.sno;
    *********************************
    7.
    *********************************
    select st.* from student st join sc s on st.sno=s.sno
    join course c on s.cno=c.cno
    join teacher t on c.tno=t.tno
    where t.tname='谌燕'
    *********************************
    8.
    *********************************
    select * from student st
    join sc a on st.sno=a.sno
    join sc b on st.sno=b.sno
    where a.cno='c002' and b.cno='c001' and a.score < b.score
    *********************************
    9.
    *********************************
    select st.*,s.score from student st
    join sc s on st.sno=s.sno
    join course c on s.cno=c.cno
    where s.score <60
    *********************************
    10.
    *********************************
    select stu.sno,stu.sname,count(sc.cno) from student stu
    left join sc on stu.sno=sc.sno
    group by stu.sno,stu.sname
    having count(sc.cno)<(select count(distinct cno)from course)
    ===================================
    select * from student where sno in
    (select sno from
            (select stu.sno,c.cno from student stu
            cross join course c
            minus
            select sno,cno from sc)
    )
    ===================================
    *********************************
    11.
    *********************************
    select st.* from student st,
    (select distinct a.sno from
    (select * from sc) a,
    (select * from sc where sc.sno='s001') b
    where a.cno=b.cno) h
    where st.sno=h.sno and st.sno<>'s001'
    *********************************
    12.
    *********************************
    select * from sc
    left join student st
    on st.sno=sc.sno
    where sc.sno<>'s001'
    and sc.cno in
    (select cno from sc
    where sno='s001')
    *********************************
    13.
    *********************************
    update sc c set score=(select avg(c.score)  from course a,teacher b
                                where a.tno=b.tno
                                and b.tname='谌燕'
                                and a.cno=c.cno
                                group by c.cno)
    where cno in(
    select cno from course a,teacher b
    where a.tno=b.tno
    and b.tname='谌燕')
    *********************************
    14.
    *********************************
    select* from sc where sno<>'s001'
    minus
    (
    select* from sc
    minus
    select * from sc where sno='s001'
    )
    *********************************
    15.
    *********************************
    delete from sc
    where sc.cno in
    (
    select cno from course c
    left join teacher t on  c.tno=t.tno
    where t.tname='谌燕'
    )
    *********************************
    16.
    *********************************
    insert into sc (sno,cno,score)
    select distinct st.sno,sc.cno,(select avg(score)from sc where cno='c002')
    from student st,sc
    where not exists
    (select * from sc where cno='c002' and sc.sno=st.sno) and sc.cno='c002';
    *********************************
    17.
    *********************************
    select cno ,max(score),min(score) from sc group by cno;
    *********************************
    18.
    *********************************
    select cno,avg(score),sum(case when score>=60 then 1 else 0 end)/count(*)
    as 及格率
    from sc group by cno
    order by avg(score) , 及格率desc
    *********************************
    19.
    *********************************
    select max(t.tno),max(t.tname),max(c.cno),max(c.cname),c.cno,avg(score) from sc , course c,teacher t
    where sc.cno=c.cno and c.tno=t.tno
    group by c.cno
    order by avg(score) desc
    *********************************
    20.
    *********************************
    select sc.cno,c.cname,
    sum(case  when score between 85 and 100 then 1 else 0 end) AS "[100-85]",
    sum(case  when score between 70 and 85 then 1 else 0 end) AS "[85-70]",
    sum(case  when score between 60 and 70 then 1 else 0 end) AS "[70-60]",
    sum(case  when score <60 then 1 else 0 end) AS "[<60]"
    from sc, course c
    where  sc.cno=c.cno
    group by sc.cno ,c.cname;
    *********************************
    21.
    *********************************
    select * from
    (select sno,cno,score,row_number()over(partition by cno order by score desc) rn from sc)
    where rn<4
    *********************************
    22.
    *********************************
    select cno,count(sno)from sc group by cno;
    *********************************
    23.
    *********************************
    select sc.sno,st.sname,count(cno) from student st
    left join sc
    on sc.sno=st.sno
    group by st.sname,sc.sno having count(cno)=1;
    *********************************
    24.
    *********************************
    select ssex,count(*)from student group by ssex;
    *********************************
    25.
    *********************************
    select * from student where sname like '张%';
    *********************************
    26.
    *********************************
    select sname,count(*)from student group by sname having count(*)>1;
    *********************************
    27.
    *********************************
    select sno,sname,sage,ssex from student t where to_char(sysdate,'yyyy')-sage =1988
    *********************************
    28.
    *********************************
    select cno,avg(score) from sc group by cno order by avg(score)asc,cno desc;
    *********************************
    29.
    *********************************
    select st.sno,st.sname,avg(score) from student st
    left join sc
    on sc.sno=st.sno
    group by st.sno,st.sname having avg(score)>85;
    *********************************
    30.
    *********************************
    select sname,score from student st,sc,course c
    where st.sno=sc.sno and sc.cno=c.cno and c.cname='Oracle' and sc.score<60
    *********************************
    31.
    *********************************
    select st.sno,st.sname,c.cname from student st,sc,course c
    where sc.sno=st.sno and sc.cno=c.cno;
    *********************************
    32.
    *********************************
    select st.sname,c.cname,sc.score from student st,sc,course c
    where sc.sno=st.sno and sc.cno=c.cno and sc.score>70
    *********************************
    33.
    *********************************
    select sc.sno,c.cname,sc.score from sc,course c
    where sc.cno=c.cno and sc.score<60 order by sc.cno desc;
    *********************************
    34.
    *********************************
    select st.sno,st.sname,sc.score from sc,student st
    where sc.sno=st.sno and cno='c001' and score>80;
    *********************************
    35.
    *********************************
    select count(distinct sno) from sc;
    *********************************
    36.
    *********************************
    select st.sname,score from student st,sc ,course c,teacher t
    where
    st.sno=sc.sno and sc.cno=c.cno and c.tno=t.tno
    and t.tname='谌燕' and sc.score=
    (select max(score)from sc where sc.cno=c.cno)
    *********************************
    37.
    *********************************
    select cno,count(sno) from sc group by cno;
    *********************************
    38.
    *********************************
    select a.* from sc a ,sc b where a.score=b.score and a.cno<>b.cno
    *********************************
    39.
    *********************************
    select * from (
    select sno,cno,score,row_number()over(partition by cno order by score desc) my_rn from sc t
    )
    where my_rn<=2
    *********************************
    40.
    *********************************
    select cno,count(sno) from sc group by cno
    having count(sno)>10
    order by count(sno) desc,cno asc;
    *********************************
    41.
    *********************************
    select sno from sc group by sno having count(cno)>1;
    ||
    select sno from sc group by sno having count(sno)>1;
    *********************************
    42.
    *********************************
    select distinct(c.cno),c.cname from course c ,sc
    where sc.cno=c.cno
    ||
    select cno,cname from course c
    where c.cno in
    (select cno from sc group by cno)
    *********************************
    43.
    *********************************
    select st.sname from student st
    where st.sno not in
    (select distinct sc.sno from sc,course c,teacher t
    where sc.cno=c.cno and c.tno=t.tno and t.tname='谌燕')
    *********************************
    44.
    *********************************
    select sno,avg(score)from sc
    where sno in
    (select sno from sc where sc.score<60
    group by sno having count(sno)>1
    ) group by sno
    *********************************
    45.
    *********************************
    select sno from sc where cno='c004' and score<90 order by score desc;
    *********************************
    46.
    *********************************
    delete from sc where sno='s002' and cno='c001';
    *********************************
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  • 原文地址:https://www.cnblogs.com/lpthread/p/3452242.html
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