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  • sql语句面试练习

    设计数据表如下

    建立数据表:

    ---------------------------------------------
    create table Student
    (
        sId int not null primary key,
        sName varchar(20) not null,
        sage int,
        sSex varchar(20)
    );
    insert into Student values 
    (01,"congcong",22,"男"),
    (02,"bingbing",23,"女"),
    (03,"fangfang",24,"女");
    
    ---------------------------------------
    create table Course
    (
        cId int not null primary key,
        cName varchar(20) not null,
        tId int not null
    );
    
    insert into Course values 
    (01,"数据结构",01),
    (02,"java基础",02),
    (03,"c++基础",02),
    (04,"操作系统",03);
    
    --------------------------------------------
    create table SC
    (
        sId int not null ,
        cId int not null ,
        score int,
        foreign key (cId) references Course(cId),
      foreign key (sId) references Student(sId)
    )
    
    insert into SC values(01,01,98);
    insert into SC values(01,02,91);
    insert into SC values(01,03,30);
    insert into SC values(02,03,80);
    insert into SC values(02,01,99);
    insert into SC values(02,02,70);
    insert into SC values(03,01,66);
    
    
    create table Teacher
    (
        tId int not null primary key,
        tName int not null
    )
    
    alter table course 
    add foreign key (tId) references Teacher(tId);
    
    alter table Teacher change tName tName varchar(20);
    insert into Teacher values(01,"张燕");
    insert into Teacher values(02,"聪姐");
    insert into Teacher values(03,"谢芳");


    建立视图,便于查选,分析

    create view view_test as
    select s.sId,s.sName,s.sage,s.sSex,sc.cId,c.cName,t.tId,t.tName,sc.score
    from Student s,sc,Course c,Teacher t
    where s.sId=sc.sId and sc.cId=c.cId and c.tId=t.tId;
    
    select * from view_test;

    查询练习:

    1、查询“01”课程比“02”课程成绩高的所有学生的学号;
    select a.sId from (select sId,score from SC where cId=01) a,(select sId,score from SC where cId=02) b
    where a.score>b.score and a.sId=b.sId;
    
    
    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;
    
    3、查询所有同学的学号、姓名、选课数、总成绩;
    select s.sId,s.sName,count(sc.cId),sum(score)
    from Student s,sc
    where s.sId=sc.sId 
    group by s.sId,s.sName;
    
    
    select s.sId,s.sName,count(sc.cId),sum(score)
    from Student s left outer join sc on s.sId=sc.sId
    group by s.sId,s.sName;
    
    4、查询姓“聪”的老师的个数;
    select count(distinct(tName))
    from Teacher
    where tName like "聪%";
    
    5、查询没学过“谢芳”老师课的同学的学号、姓名;
    select Student.sId,Student.sName
    from Student
    where sId not in (select sc.sId from sc,Course,Teacher where sc.cId=Course.cId
    and Course.tId=Teacher.tId and Teacher.tName="谢芳" );
    
    6、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名; 
    select sId,sName from Student 
    where sId in(select sc.sId from sc,Student where cId=01 and sc.sId=Student.sId and  sc.sId
    in (select sc.sId from sc,Student where cId=02 and sc.sId=Student.sId ));
    
    select Student.sId,Student.sName
    from Student,sc 
    where Student.sId=sc.sId and sc.cId=01 and
    exists(select * from sc as sc_2 where sc_2.sId=sc.sId and sc_2.cId=02);
    
    
    7、查询学过“聪姐”老师所教的同学的学号、姓名;
    
    select distinct s.sId,s.sName
    from Student s,sc,Course c,Teacher t
    where t.tName="聪姐" and c.tId=t.tId and c.cId=sc.cId and sc.sId=s.sId;
    
    8、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名;
    
    select result.sId,result.sName 
    from (select s.sId,s.sName,sc.score,(select score from sc sc_2 where sc_2.sId=s.sId and sc_2.cId=02) score2 
    from Student s,sc where s.sId=sc.sId and sc.cId=01)result where score2<score;
    
    9、查询所有课程成绩小于60分的同学的学号、姓名
    
    select result.sId,result.sName
    from (select s.sId,s.sName,sc.score from Student s,sc
    where s.sId=sc.sId and sc.score>60) result;
    
    10、查询没有学全所有课的同学的学号、姓名;
    
    select s.sId,s.sName
    from Student s,sc
    where s.sId=sc.sId 
    group by s.sId,s.sName 
    having count(sc.cId)<(select count(cId) from Course);
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  • 原文地址:https://www.cnblogs.com/huangcongcong/p/4004811.html
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