• 50SQL练习


    Student(Sid,Sname,Sage,Ssex) 学生表

    Course(Cid,Cname,Tid) 课程表
    SC(Sid,Cid,score) 成绩表
    Teacher(Tid,Tname) 教师表
    练习内容:
    1.查询“某1”课程比“某2”课程成绩高的所有学生的学号;
    SELECT a.sid FROM (SELECT sid,score FROM SC WHERE cid=1) a,(SELECT sid,score FROM SC WHERE cid=3) b WHERE a.score>b.score AND a.sid=b.sid;
    此题知识点,嵌套查询和给查出来的表起别名
    2.查询平均成绩大于60分的同学的学号和平均成绩;
    SELECT sid,avg(score)  FROM sc  GROUP BY sid having avg(score) >60;
    此题知识点,GROUP BY 语句用于结合合计函数,根据一个或多个列对结果集进行分组。group by后面不能接where,having代替了where
    3.查询所有同学的学号、姓名、选课数、总成绩
    SELECT Student.sid,Student.Sname,count(SC.cid),sum(score)FROM Student left Outer JOIN SC on Student.sid=SC.cid GROUP BY Student.sid,Sname
    4.查询姓“李”的老师的个数;
    select count(teacher.tid)from teacher where teacher.tname like'李%'
    5.查询没学过“叶平”老师课的同学的学号、姓名;
    SELECT Student.sid,Student.Sname FROM Student WHERE sid not in (SELECT distinct( SC.sid) FROM SC,Course,Teacher WHERE  SC.cid=Course.cid AND Teacher.id=Course.tid AND Teacher.Tname='叶平');
     此题知识点,distinct是去重的作用
    6.查询学过“```”并且也学过编号“```”课程的同学的学号、姓名;
    select a.SID,a.SNAME from (select student.SNAME,student.SID from student,course,sc where cname='c++'and sc.sid=student.sid and sc.cid=course.cid) a,
    (select student.SNAME,student.SID from student,course,sc where cname='english'and sc.sid=student.sid and sc.cid=course.cid) b where a.sid=b.sid;
    标准答案(但是好像不好使)SELECT Student.S#,Student.Sname FROM Student,SC WHERE Student.S#=SC.S# AND SC.C#='001'and exists( SELECT * FROM SC as SC_2 WHERE SC_2.S#=SC.S# AND SC_2.C#='002');  
    此题知识点,exists是在集合里找数据,as就是起别名
    7.查询学过“叶平”老师所教的所有课的同学的学号、姓名;
    select a.sid,a.sname from (select student.sid,student.sname from student,teacher,course,sc 
    where teacher.TNAME='杨巍巍' and teacher.tid=course.tid and course.cid=sc.cid and student.sid=sc.sid) a
    标准答案:SELECT sid,Sname FROM Student WHERE sid in (SELECT sid FROM SC ,Course ,Teacher WHERE SC.cid=Course.cid AND Teacher.tid=Course.tid AND Teacher.Tname='杨巍巍' GROUP BY sid having count(SC.cid)=(SELECT count(cid) FROM Course,Teacher  WHERE Teacher.tid=Course.tid AND Tname='杨巍巍'))
    8.查询课程编号“”的成绩比课程编号“”课程低的所有同学的学号、姓名;
    select a.sid,a.sname from(select student.SID,student.sname,sc.SCORE  from student,sc where student.sid=sc.sid and sc.cid=1) a,
    (select student.SID,student.sname,sc.score from student,sc where student.sid=sc.sid and sc.cid=2) b where a.score<b.score and a.sid=b.sid
    标准答案:SELECT sid,Sname FROM (SELECT Student.sid,Student.Sname,score ,
    (SELECT score FROM SC SC_2 WHERE SC_2.sid=Student.sid AND SC_2.cid=1) score2 FROM Student,SC
    WHERE Student.sid=SC.sid AND cid=1) S_2 WHERE score2 <score;
    9.查询所有课程成绩小于分的同学的学号、姓名;
    SELECT sid,Sname FROM Student WHERE sid not in (SELECT Student.sid FROM Student,SC WHERE Student.sid=SC.sid AND score>60); 
    此题知识点,先查出大于60分的,然后not in 就是小于60分的了
    10.查询没有学全所有课的同学的学号、姓名;
    SELECT Student.sid,Student.Sname  FROM Student,SC  
    WHERE Student.sid=SC.sid GROUP BY  Student.sid,Student.Sname having count(cid) <(SELECT count(cid) FROM Course); 
    11.查询至少有一门课与学号为“”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“”同学所有一门课的其他同学学号和姓名;
    SELECT student.sid,student.Sname FROM Student,SC WHERE Student.sid=SC.sid AND cid in (SELECT cid FROM SC WHERE sid=1)
    此题知识点,SELECT sid,Sname FROM Student,SC WHERE Student.sid=SC.sid AND cid in (SELECT cid FROM SC WHERE sid=1)这样写是错误的,因为from后面是两个表,不能明确是哪个表里面的sid和sname所以错误提示是“未明确定义列”
    13.把“SC”表中“叶平”老师教的课的成绩都更改为此课程的平均成绩;
    update sc set score=(select avg(score) from sc,course,teacher where course.cid=sc.cid and course.tid=teacher.tid and teacher.tname='杨巍巍')
    14.查询和“”号的同学学习的课程完全相同的其他同学学号和姓名;
    SELECT sid FROM SC WHERE cid in (SELECT cid FROM SC WHERE sid=6) GROUP BY sid having count(*)=(SELECT count(*) FROM SC WHERE sid=6); 
    此题知识点,用数量来判断 
    15.删除学习“叶平”老师课的SC表记录; 
    delete from sc s where s.cid in (select c.cid from teacher t,course c where t.tid = c.tid and tname='李子')
    此题知识点,嵌套查询可以分布考虑,先查出李子老师都交了什么课的id,然后再删除那些id的值
    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“”课程的同学学号、课程的平均成绩;
    Insert into SC SELECT sid,2,(SELECT avg(score) FROM SC WHERE cid=2) FROM Student WHERE sid not in (SELECT sid FROM SC WHERE cid=2); 
    17.按平均成绩从高到低显示所有学生的“数据库”、“企业管理”、“英语”三门的课程成绩,按如下形式显示:学生ID,,数据库,企业管理,英语,有效课程数,有效平均分;(没做出来)
    18.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;
    select cid as 课程号,max(score)as 最高分,min(score) as 最低分 from sc group by cid
    标准答案(但是运行不好使)SELECT L.cid As 课程ID,L.score AS 最高分,R.score AS 最低分 
    FROM SC L ,SC AS R  
    WHERE L.cid = R.cid AND  
    L.score = (SELECT MAX(IL.score)  
    FROM SC AS IL,Student AS IM  
    WHERE L.cid = IL.cid AND IM.sid=IL.sid  
    GROUP BY IL.cid)  
    AND  R.Score = (SELECT MIN(IR.score) FROM SC AS IR WHERE R.cid = IR.cid  GROUP BY IR.cid ); 
    19.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
    26.查询每门课程被选修的学生数
    select sc.cid,count(sc.sid) from sc,course where sc.cid=course.cid group by sc.cid
    27.查询出只选修了一门课程的全部学生的学号和姓名
    SELECT SC.sid,Student.Sname,count(cid) AS 选课数 FROM SC ,Student  
    WHERE SC.sid=Student.sid GROUP BY SC.sid ,Student.Sname having count(cid)=1;
    32.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
    SELECT Cid,Avg(score) FROM SC GROUP BY cid ORDER BY Avg(score),cid DESC ;
    37.查询不及格的课程,并按课程号从大到小排列 
    SELECT cid,sid FROM sc WHERE score <60 ORDER BY cid 
    38.查询课程编号为且课程成绩在分以上的学生的学号和姓名;
    select student.sid,student.sname from sc,student where sc.cid=1 and sc.score>60 and sc.sid=student.sid
    40.查询选修“叶平”老师所授课程的学生中,成绩最高的学生姓名及其成绩
    select student.sname,sc.score from sc,student,teacher,course c where teacher.tname='李子'
    and teacher.tid=c.tid and c.cid=sc.cid and sc.sid=student.sid and sc.score=(select max(score)from sc where sc.cid=c.cid)
    41.查询各个课程及相应的选修人数
    select sc.cid ,count(sc.sid)from sc,student where sc.sid=student.sid group by sc.cid 
    43.查询每门功成绩最好的前两名
     
    44.统计每门课程的学生选修人数(超过人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列,查询结果按人数降序排列,若人数相同,按课程号升序排列
    select sc.cid,count(sc.cid)from sc,course where sc.cid=course.cid group by sc.cid  order by sc.cid desc
    45.检索至少选修两门课程的学生学号
    SELECT sid FROM  sc group  by  sid having  count(*)  >  =  2  
     
    rownum的用法
    查询所有成绩第二名到第四名的成绩
    select * from (select rownum p,t.score from(SELECT s.score score FROM sc s ORDER BY score desc)t )tt where tt.p>1 and tt.p<5
     
    47.查询没学过“叶平”老师讲授的任一门课程的学生姓名
    select distinct sid from sc where sid not in(select sc.sid from sc,course,teacher where sc.cid=course.cid and course.tid=teacher.tid and 
    teacher.tname='杨巍巍')
    48.查询两门以上不及格课程的同学的学号及其平均成绩
    49.检索“”课程分数小于,按分数降序排列的同学学号
    select sc.sid from sc,course where sc.cid=course.cid and course.cname='java' and sc.score<90
    50.删除“”同学的“”课程的成绩 
    delete from sc where sid=1 and cid=1
  • 相关阅读:
    算法
    什么叫「人的格局」?是否有必要培养大的格局或怎么培养?
    套接字
    信号量 P V测试详解
    Emacs快速入门
    Emacs 从入门到精通
    Linux多进程之间的文件锁
    mysql 登录报错:ERROR 1045 (28000)
    类职责协作模型
    .Netcore使用Session
  • 原文地址:https://www.cnblogs.com/yang-taiyue/p/6849779.html
Copyright © 2020-2023  润新知