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  • mysql经典查询

    建立数据库

    1.建立一个数据库 
    create database work;

    2.进入数据库work 
    use work;

    3.数据库默认编码可能不支持中文,可以在这里设置下 
    set names gbk;

    4.建立student表 
    属性有:编号:id (主键,自动增长),姓名:sname,出生年月:sage,性别:ssex(枚举) 
    create table student(sid int primary key auto_increment, 
    sname varchar(20), 
    sage date, 
    ssex enum(‘男’,’女’));

    5.第二个课程表中使用了外键教师标号,因而需要先建立教师表 
    create table teacher(tid int primary key auto_increment, 
    tname varchar(20));

    6.建立课程表 
    create table course(cid int primary key auto_increment, 
    cname varchar(20), 
    tid int, 
    foreign key(tid) references teacher(tid));

    7.建立成绩表 
    create table sc(sid int, 
    cid int, 
    score int);

    8.show tables; //可查看建立的四个表格

    9.插入数据,因为里面有主键链接,表格插入数据也要有顺序(注意题目图片上都是字节引号,应该为int,不要单引号)

    a,先给student表插入数据
    insert into student values(1,'赵雷','1990-01-01','男'),
        (2,'钱电','1990-12-21','男'),
        (3,'孙风','1990-05-20','男'),
        (4,'李云','1990-08-06','男'),
        (5,'周梅','1991-12-01','女'),
        (6,'吴兰','1992-03-01','女'),
        (7,'郑竹','1989-07-01','女'),
        (8,'王菊','1990-01-20','女');
    
    b, 给teacher表插入数据,这里不可以先给course表插入数据,因为course表外链接到teacher的主键
    insert into teacher values(1,'张三'),
            (2,'李四'),
            (3,'王五');
    
    c, 给course表插入数据
    insert into course values(1,'语文',2),
                (2,'语文',1),
                (3,'语文',3);
    
    d, 最后给sc表插入数据(题目图片少了第1个学生成绩,在这加上 1,1,90;  1,2,80;  1,3,90)
    insert into sc values(1,1,90),
                (1,2,80),
                (1,3,90),
                (2,1,70),
                (2,2,60),
                (2,3,80),
                (3,1,80),
                (3,2,80),
                (3,3,80),
                (4,1,50),
                (4,2,30),
                (4,3,20),
                (5,1,76),
                (5,2,87),
                (6,1,31),
                (6,3,34),
                (7,2,89),
                (7,3,98);
    
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    ———————–数据库建立完成—————————————

    1、查询”01”课程比”02”课程成绩高的学生的信息及课程分数 
    select s.sid,s.sname,s.sage,s.ssex,sc1.score,sc2.score from student s,sc sc1,sc sc2 where sc1.cid=1 and sc2.cid=2 and sc1.score>sc2.score and sc1.sid=sc2.sid and s.sid=sc1.sid;

    2、查询同时存在”01”课程和”02”课程的情况 
    select * from course c1,course c2;

    3.查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩 
    select s.sid,s.sname,avg(score) from student s,sc group by s.sid having avg(score)>60;

    4、查询名字中含有”风”字的学生信息 
    select * from student where sname like ‘%风%’;

    5、查询课程名称为”数学”,且分数低于60的学生姓名和分数 
    select s.sname,score from student s,sc where s.sid=sc.sid and cid=2 and score<60;

    6、查询所有学生的课程及分数情况; 
    select cname,score from sc,course where sc.cid=course.cid;

    7、查询没学过”张三”老师授课的同学的信息 
    select s.* from student s where s.sid not in(select sc1.sid from sc sc1,course c,teacher t where t.tid=c.tid and sc1.cid=c.cid and t.tname=’张三’);

    8.查询学过”张三”老师授课的同学的信息 
    select s.* from student s ,sc sc1,course c,teacher t where s.sid=sc1.sid and sc1.cid=c.cid and c.tid=t.tid and t.tname=’张三’;

    9、查询学过编号为”01”并且也学过编号为”02”的课程的同学的信息 
    student(sid) sc(sid cid tid) sc2(sid cid tid) course(cid tid cname) 
    select s.* from student s,sc sc1,sc sc2 where s.sid=sc1.sid and sc1.sid=sc2.sid and sc1.cid=1 and sc2.cid=2;

    10、查询学过编号为”01”但是没有学过编号为”02”的课程的同学的信息 
    select distinct s.* from student s,sc sc1,sc sc2,sc sc3 where s.sid=sc1.sid and sc1.sid=sc2.sid and sc1.cid=1 and sc2.cid!=2;

    11、查询没有学全所有课程的同学的信息 
    select s.* from student s where s.sid not in(select sc1.sid from sc sc1,sc sc2,sc sc3 where sc1.cid=1 and sc2.cid=2 and sc3.cid =3 and sc1.sid=sc2.sid and sc1.sid=sc3.sid) group by s.sid;

    12、查询至少有一门课与学号为”01”的同学所学相同的同学的信息 
    select distinct s.* from student s,sc sc1 where s.sid=sc1.sid and sc1.cid in(select cid from sc where sid=1) and s.sid<> 1;

    13、查询和”01”号的同学学习的课程完全相同的其他同学的信息 
    select s.* from student s where s.sid in(select distinct sc.sid from sc where sid<>1 and sc.cid in(select distinct cid from sc where sid=1)group by sc.sid having count(1)=(select count(1) from sc where s.sid=1));

    14、查询没学过”张三”老师讲授的任一门课程的学生姓名 
    select s.* from student s where s.sid not in(select sc1.sid from sc sc1,course c,teacher t where sc1.cid=c.cid and c.tid=t.tid and t.tname=’张三’);

    15、查询出只有两门课程的全部学生的学号和姓名 
    select s.* from student s,sc group by sc.sid having count(sc.sid)=2 and s.sid=sc.sid;

    16、查询1990年出生的学生名单(注:Student表中Sage列的类型是datetime) 
    select * from student where sage>=’1900-01-01’ and sage<=’1900-12-31’; 
    select s.* from student s where s.sage like ‘1900-%’;(方法2)

    17、查询每门课程的平均成绩,结果按平均成绩降序排列,平均成绩相同时,按课程编号升序排列 
    select sc.cid,avg(score) from sc group by sc.cid order by avg(score) DESC , sc.cid;

    18、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; 
    select s.sname,c.cname,score from student s,sc,course c where s.sid=sc.sid and sc.cid=c.cid and score>70;

    19、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩 
    select s.sname,avg(score) from sc,student s where s.sid=sc.sid group by sc.sid having avg(score)>=85;

    20、查询不及格的课程 
    select s.sname,c.cname,score from student s,sc,course c where s.sid=sc.sid and sc.cid=c.cid and score<60;

    21、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名; 
    select s.sid,s.sname from student s,sc where sc.sid=s.sid and sc.cid=1 and score>80;

    22、求每门课程的学生人数 
    select cid,count(sid) from sc group by sc.cid;

    23、统计每门课程的学生选修人数(超过5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 
    select cid,count(sid) from sc group by cid having count(sid)>5 order by count(sid),cid ASC;

    24、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩 
    select s1.sid,s2.sid,sc1.cid,sc1.score,sc2.score from student s1,student s2,sc sc1,sc sc2 where s1.sid!=s2.sid and s1.sid=sc1.sid and s2.sid=sc2.sid and sc1.cid!=sc2.cid and sc1.score=sc2.score;

    25、检索至少选修两门课程的学生学号 
    select sid from sc group by sid having count(cid)>=2;

    26、查询选修了全部课程的学生信息 
    select s.* from sc,student s where s.sid=sc.sid group by sid having count 
    (cid)=3;

    27、查询各学生的年龄 
    select s.sname,(TO_DAYS(‘2017-09-07’)-TO_DAYS(s.sage))/365 as age from student s;

    28、查询本月过生日的学生 
    select s.sname from student s where s.sage like ‘_____07%’;

    29、查询下月过生日的学生 
    select s.sname from student s where s.sage like ‘_____08%’;

    30、查询学全所有课程的同学的信息 
    select s.* from student s,sc sc1,sc sc2,sc sc3 where sc1.cid=1 and sc2.cid=2 and sc3.cid=3 and sc1.sid=sc2.sid and sc1.sid=sc3.cid and s.sid =sc1.sid group by s.sid;

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  • 原文地址:https://www.cnblogs.com/hanxue112253/p/9657871.html
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