• mysql 查询的练习与提高


    -- 创建数据库school
    create database school;
    
    -- 选择进入school数据库
    use school;
    
    
    -- ------------建表导数-------------
    -- 创建stu
    create table stu(
    s_id varchar(10) primary key,
    s_name varchar(10) not null,
    s_birth date,
    s_sex varchar(10));
    
    -- 导入数据
    insert into stu values
    ('01' , '赵雷' , '1990-01-01' , '男'),
    ('02' , '钱电' , '1990-12-21' , '男'),
    ('03' , '孙风' , '1990-05-20' , '男'),
    ('04' , '李云' , '1990-08-06' , '男'),
    ('05' , '周梅' , '1991-12-01' , '女'),
    ('06' , '吴兰' , '1992-03-01' , '女'),
    ('07' , '郑竹' , '1992-04-21' , '女'),
    ('08' , '王菊' , '1990-01-20' , '女');
    
    select * from stu; -- 检查数据
    select count(*) from stu; -- 检查总行数
    
    
    -- 创建co
    create table co(
    c_id varchar(10) primary key,
    c_name varchar(10),
    t_id varchar(10));
    
    -- 导入数据
    insert into co values
    ('01' , '语文' , '02'),
    ('02' , '数学' , '01'),
    ('03' , '英语' , '03');
    
    select * from co; -- 检查数据
    select count(*) from co; -- 检查总行数
    
    
    -- 创建te
    create table te(
    t_id varchar(10) primary key,
    t_name varchar(10));
    
    -- 导入数据
    insert into te values
    ('01' , '张三'),
    ('02' , '李四'),
    ('03' , '王五');
    
    select * from te; -- 检查数据
    select count(*) from te; -- 检查总行数
    
    
    -- 创建sc
    create table sc(
    s_id varchar(10),
    c_id varchar(10),
    score int);
    
    -- 导入数据
    insert into sc values
    ('01' , '01' , 80),
    ('01' , '02' , 90),
    ('01' , '03' , 99),
    ('02' , '01' , 70),
    ('02' , '02' , 60),
    ('02' , '03' , 80),
    ('03' , '01' , 80),
    ('03' , '02' , 80),
    ('03' , '03' , 80),
    ('04' , '01' , 50),
    ('04' , '02' , 30),
    ('04' , '03' , 20),
    ('05' , '01' , 76),
    ('05' , '02' , 87),
    ('06' , '01' , 31),
    ('06' , '03' , 34),
    ('07' , '02' , 89),
    ('07' , '03' , 98);
    
    select * from sc; -- 检查数据
    select count(*) from sc; -- 检查总行数
    

      练习题

    -- ----------------查询提高------------------
    -- 1、查询考试成绩不及格的学生信息
    select s_id 
    from sc
    where score <60;
    
    select *
    from stu
    where s_id in 
    (select s_id 
    from sc
    where score <60);
    
    select distinct stu.*
    from sc left join stu on stu.s_id=sc.s_id
    where score < 60;
    
    #笛卡儿积  不能有非匹配项
    select distinct stu.*
    from stu , sc
    where stu.s_id = sc.s_id and score <60;
    
    
    
    -- 2、查询姓‘李’的老师的数量   ##模糊查询
    select count(t_id)
    from te
    where t_name like '李%';
    
    
    -- 3、查询没有出现过考试成绩不及格的学生:s_id,c_name,score
    ##1、考试成绩大于60 的学生
    select distinct s_id 
    from  sc
    where score<60;
    
    select stu.s_id,stu.s_name,co.c_name,score
    from stu left join sc on stu.s_id = sc.s_id left join co on sc.c_id = co.c_id
    where stu.s_id not in (
    select distinct s_id 
    from  sc
    where score<60);
    
    
    -- 4、查询平均成绩大于70分的课程
    #第一种方法
    select c_name, avg(score) ,co.c_id
    from sc left join co on sc.c_id = co.c_id
    group by sc.c_id
    having avg(score)>70;
    
    #第二种方法
    
    select c_id 
    from sc
    group by c_id
    having avg(score)>70;
    
    select c_name ,c_id
    from co 
    where c_id in
    (
    select c_id 
    from sc
    group by c_id
    having avg(score)>70);
    
    -- 5、查询平均成绩大于60分的学生:s_id,s_name,平均成绩
    select stu.s_id , s_name , avg(score)
    from sc left join stu on stu.s_id = sc.s_id
    group by sc.s_id
    having avg(score)>60;
    
    ##第二种方法
    select s_id,avg(score)
    from sc
    group by s_id
    having avg(score)>60; 
    
    select s_id , s_name 
    from stu
    where s_id in
    (select s_id
    from sc
    group by s_id
    having avg(score)>60);
    -- 6、查询至少选修了两门课程的学生:s_id,s_name,选课门数
    select sc.s_id,s_name,count(sc.s_id)
    from sc left join stu on sc.s_id = stu.s_id
    group by sc.s_id
    having count(sc.s_id) >=2;
    
    ##第二种方法
    select s_id ,count(c_id)
    from sc 
    group by s_id
    having count(c_id) >= 2;
    
    select s.*,s_name
    from (select s_id ,count(c_id)
    from sc 
    group by s_id
    having count(c_id) >= 2) s left join stu on s.s_id=stu.s_id;
    
    -- 7、查询选修了全部课程的学生信息   ##思路:先查询出总共的课程门数有多少,然后连接学生表和成绩表,通过学生id分组,当成绩表中的选修门数等于所有的课程数时输出
    #查找课程门数 
     
    select count(c_id) from co;
    
    select stu.*,count(sc.c_id) 选修门数
    from sc left join stu on sc.s_id = stu.s_id
    group by s_id
    having 选修门数=(select count(c_id) from co);
    
    
    
    -- 8、查询每门课程被选修的学生数   ##课程表中的
    select co.c_id,c_name,count(*) 选修人数
    from co left join sc on co.c_id = sc.c_id
    group by co.c_id;
    
    
    
    
    
    -- 9、查询90后学生的男女人数
    ##第一种方法
    select sum(datediff(s_birth,'1989-12-12')>0) 90后人数,s_sex   #90后的人数
    from stu
    group by s_sex;
    
    ##第二种方法
    select s_sex,count(s_id) 人数
    from stu
    where year(s_birth)>=1990
    group by s_sex;
    
    -- 10、查询至少有3门课程及格的学生
    
    
    select sc.s_id ,s_name,count(score>60) 
    from sc left join stu on sc.s_id = stu.s_id
    where score >=60
    group by sc.s_id
    having count(score>60)>=3;
    
    
    ##查询每位同学的及格门数
    select s_id , count(c_id)  及格门数
    from sc
    where score >= 60
    group by s_id
    having 及格门数>=3;
    
    #子查询
    select s.s_id,s_name ,及格门数
    from (
    select s_id , count(c_id)  及格门数
    from sc
    where score >= 60
    group by s_id
    having 及格门数>=3) s 
    left join stu on s.s_id = stu.s_id;
    
    -- 11、查询所有学生的s_id,s_name,选课门数,所选课程总成绩
    select stu.s_id,s_name,count(c_id),sum(score)
    from stu left join sc on stu.s_id = sc.s_id
    group by stu.s_id;
    
    
    -- 12、查询至少有两门成绩不及格的学生:s_id,s_name,不及格门数,平均成绩
    #第一种方法
    select sc.s_id,s_name,count(score<60),avg(score)
    from sc left join stu on sc.s_id= stu.s_id
    where score <60
    group by sc.s_id
    having count(score<60)>=2;
    #第二种方法
    select sc.s_id,s_name,count(c_id) 不及格门数,avg(score)
    from sc left join stu on stu.s_id= sc.s_id
    group by sc.s_id
    having sum(score<60)>=2;
    
    select sum(score <60) from sc;
    -- 13、学过"张三"老师课程的学生信息   #四表连接顺序, 教师,课程,成绩,学生
    
    #方法一
    select *
    from te left join co on te.t_id = co.t_id 
    left join sc on co.c_id = sc.c_id 
    left join stu on sc.s_id = stu.s_id
    where t_name = '张三';
     
     #方法二  子查询
    select *
    from stu
    where s_id in
    (select s_id
    from te left join co on te.t_id = co.t_id left join sc on co.c_id = sc.c_id
    where t_name = '张三');
    
    
    -- 14、查询所有学生不同课程的成绩及平均成绩:s_id,语文,数学,英语,平均成绩
    #第一种方法
    select s_id,sum(case when c_id='01' then score else 0 end) 语文,
    			sum(case when c_id='02' then score else 0 end) 数学,
    			sum(case when c_id='03' then score else 0 end) 英语,
    			avg(score) 平均成绩
    from sc
    group by s_id;
    
    
    select stu.s_id,sum(case when c_id='01' then score else 0 end) 语文,
    			sum(case when c_id='02' then score else 0 end) 数学,
    			sum(case when c_id='03' then score else 0 end) 英语,
    			avg(score) 平均成绩
    from stu left join sc on stu.s_id=sc.s_id
    group by stu.s_id;
    
    ##第二种方法
    select s_id,sum(if( c_id='01',score ,0)) 语文,
    			sum(if( c_id='02',score ,0)) 数学,
    			sum(if( c_id='03',score ,0)) 英语,
    			avg(score) 平均成绩
    from sc
    group by s_id;
    -- 15、查询不同老师所授不同课程的平均成绩
    select t_name,avg(score),c_name
    from te left join co on te.t_id = co.t_id left join sc on co.c_id = sc.c_id
    group by te.t_id,co.c_id;
    
    
    
    -- 16、查询和01号同学所学课程完全一样的学生信息
    select group_concat(c_id) from sc group by s_id having s_id = '01'; #找出01所选修的课程
    
    
    select c_id
    from sc
    where s_id='01';
    
    select stu.s_id,group_concat(c_id order by c_id) 选修课程
    from stu left join sc on sc.s_id= stu.s_id
    group by stu.s_id
    having 选修课程 = (select group_concat(c_id order by c_id) from sc group by s_id having s_id = '01');
    
    
    
    -- 17、查询至少有一门课与01号同学所学课程一样的学生信息
    select c_id
    from sc
    where s_id='01';
    
    select distinct stu.*
    from stu left join sc on sc.s_id= stu.s_id
    where c_id in  (select c_id
    from sc
    where s_id='01');
    
    
    
    
    -- 18、查询选修了01号课程并且也选修了02号课程的学生信息
    
    ##第一种方法
    select s_id,count(c_id)
    from sc
    where c_id in (01,02)
    group by s_id
    having count(c_id)=2;
    
    
    select stu.*
    from sc left join stu on sc.s_id= stu.s_id
    where c_id in (01,02)
    group by sc.s_id
    having count(c_id)=2;
    
    ##第二种方法
    select distinct stu.*
    from sc left join stu on sc.s_id= stu.s_id
    where stu.s_id in (select s_id from sc where c_id='01') and 
    		stu.s_id in(select s_id from sc where c_id='02');
    
    ##第三种方法
    select stu.*,group_concat(c_id order by c_id) 选修课程
    from sc left join stu on sc.s_id= stu.s_id
    group by stu.s_id
    having 选修课程 like '01,02%';
    
    #第四种方法
    select stu.*
    from sc left join stu on sc.s_id= stu.s_id
    group by stu.s_id
    having sum(c_id=01)+sum(c_id=02)=2;
    having sum(c_id =(01,02))=2;
    -- 19、查询01号课程比02号课程分数高的学生信息及成绩
    #第一步
    select s_id,score from sc where c_id = '01';
    select s_id,score from sc where c_id = '02';
    
    #第二步
    select *
    from stu left join (select s_id,score from sc where c_id = '01') s1 on stu.s_id=s1.s_id left join
    (select s_id,score from sc where c_id = '02') s2 on stu.s_id=s2.s_id
    where s1.score>s2.score;
    
    
    -- 20、查询选修了01号课程但是没有选修02号课程的学生信息
    select distinct stu.*
    from sc left join stu on sc.s_id= stu.s_id
    where stu.s_id in (select s_id from sc where c_id='01') and 
    		stu.s_id not in(select s_id from sc where c_id='02');
    
    #第二种方法
    select stu.*
    from sc left join stu on sc.s_id= stu.s_id
    group by stu.s_id
    having sum(c_id=01)=1 and sum(c_id=02)=0;
    
    
    
    
    
    
    
    
    
    -- 21、查询选修'张三'老师所授课程的学生中,分数最高的学生信息及成绩
    select stu.*,score
    from te left join co on te.t_id = co.t_id 
    left join sc on co.c_id = sc.c_id 
    left join stu on sc.s_id = stu.s_id
    where t_name = '张三'
    order by score desc
    limit 1;
    
    select stu.*,max(score)
    from te left join co on te.t_id = co.t_id 
    left join sc on co.c_id = sc.c_id 
    left join stu on sc.s_id = stu.s_id
    where t_name = '张三';
    
    
    -- 22、查询各科成绩最高分,最低分,平均分,合格率:cname
    select c_name ,max(score),min(score),avg(score),sum(score>=60)/count(co.c_id) 合格率
    from co left join sc on sc.c_id = co.c_id
    group by co.c_id;
    
    ##考试成绩合格的人数
    select sum(score >=60) from sc;
    -- 23、统计各科成绩各分数段人数
    select co.c_id ,c_name,
    sum(score >=85) '[100,85]',
    sum(score >=70 and score <85) '[70,85]',
    sum(score >=60 and score <70 ) '[60,70]',
    sum( score <60 ) '[0,60]'
    from co left join sc on co.c_id = sc.c_id
    group by co.c_id;
    
    
    select c_id , sum(case when score >=85 then 1 else 0 end) '[100,85]',
    sum(case when score between 70 and 85 then 1 else 0 end) '[70,85]',
    sum(case when score between 60 and 70 then 1 else 0 end) '[60,70]',
    sum(case when score <60 then 1 else 0 end) '[0,60]'
    from sc
    group by c_id;
    
    
    
    -- 24、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 *
    from sc
    group by s_id
    having max(score)<70;
    
    select s_name,c_name,score
    from stu 
    left join sc on sc.s_id= stu.s_id 
    left join co on sc.c_id = co.c_id
    	where stu.s_id not in (
    	select s_id
    	from sc
    	group by s_id
    	having max(score)<70);
    
    
    
    /*
    select s_name,c_name,score
    from sc left join stu on sc.s_id= stu.s_id left join co on sc.c_id = co.c_id
    where score>70
    group by score;*/
    -- 25、按各科成绩进行排序,并显示排名;s_id,c_id,score,排名
    
    use school;
    set @ss:=0;
    
    select a.*,@ss:=@ss+1 as 排名
    from (select c_id,s_id,score
    from sc
    group by c_id,s_id
    order by c_id,score desc) a;
    
    select c_id,s_id,score
    from sc
    group by c_id,s_id
    order by c_id,score desc;
    
    
    
    
    select @ss;
    

      

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  • 原文地址:https://www.cnblogs.com/manjianlei/p/11284676.html
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