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oracle查询练习

oracle查询练习

作者: 程序员bobo | 来源:发表于2018-12-10 05:39 被阅读21次

    create table student(
    sno varchar2(10) primary key,
    sname varchar2(20),
    sage number(2),
    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;

    练习:
    注意:以下练习中的数据是根据初始化到数据库中的数据来写的SQL 语句,请大家务必注意。

    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:


    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)



    select sno,avg(score) from sc group by sno having avg(score)>60;



    select a.*,s.sname from (select sno,sum(score),count(cno) from sc group by sno) a ,student s where a.sno=s.sno



    select count(*) from teacher where tname like '刘%';



    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')



    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;



    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'



    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



    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



    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)
    )
    ===================================



    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'



    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')



    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')



    select student.sno,student.sname from student,sc
    where student.sno=sc.sno
    and sc.cno in (select sc.cno from sc where sc.sno='s001') and sc.sno<>'s001'
    group by student.sno,student.sname
    having count()=(select count() from sc where sno='s001');



    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'
    )



    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';



    select cno ,max(score),min(score) from sc group by cno;



    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



    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



    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;



    select * from
    (select sno,cno,score,row_number()over(partition by cno order by score desc) rn from sc)
    where rn<4



    select cno,count(sno)from sc group by cno;



    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;



    select ssex,count(*)from student group by ssex;



    select * from student where sname like '张%';



    select sname,count()from student group by sname having count()>1;



    select sno,sname,sage,ssex from student t where to_char(sysdate,'yyyy')-sage =1988



    select cno,avg(score) from sc group by cno order by avg(score)asc,cno desc;



    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;



    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



    select st.sno,st.sname,c.cname from student st,sc,course c
    where sc.sno=st.sno and sc.cno=c.cno;



    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



    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;



    select st.sno,st.sname,sc.score from sc,student st
    where sc.sno=st.sno and cno='c001' and score>80;



    select count(distinct sno) from sc;



    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)



    select cno,count(sno) from sc group by cno;



    select a.* from sc a ,sc b where a.score=b.score and a.cno<>b.cno



    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



    select cno,count(sno) from sc group by cno
    having count(sno)>10
    order by count(sno) desc,cno asc;



    select sno from sc group by sno having count(cno)>1;
    ||
    select sno from sc group by sno having count(sno)>1;



    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)



    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')



    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



    select sno from sc where cno='c004' and score<90 order by score desc;



    delete from sc where sno='s002' and cno='c001';


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