SQL语句练习50题(MySQL版)

作者: Java架构师CAT | 来源:发表于2019-08-07 17:58 被阅读1次

    表名和字段

    –1.学生表

    Student(s_id,s_name,s_birth,s_sex) --学生编号,学生姓名, 出生年月,学生性别

    –2.课程表

    Course(c_id,c_name,t_id) – --课程编号, 课程名称, 教师编号

    –3.教师表

    Teacher(t_id,t_name) --教师编号,教师姓名

    –4.成绩表

    Score(s_id,c_id,s_score) --学生编号,课程编号,分数

    测试数据

    --建表

    --学生表

    CREATE TABLE `Student`(

      `s_id` VARCHAR(20),

      `s_name` VARCHAR(20) NOT NULL DEFAULT '',

      `s_birth` VARCHAR(20) NOT NULL DEFAULT '',

      `s_sex` VARCHAR(10) NOT NULL DEFAULT '',

      PRIMARY KEY(`s_id`)

    );

    --课程表

    CREATE TABLE `Course`(

      `c_id`  VARCHAR(20),

      `c_name` VARCHAR(20) NOT NULL DEFAULT '',

      `t_id` VARCHAR(20) NOT NULL,

      PRIMARY KEY(`c_id`)

    );

    --教师表

    CREATE TABLE `Teacher`(

      `t_id` VARCHAR(20),

      `t_name` VARCHAR(20) NOT NULL DEFAULT '',

      PRIMARY KEY(`t_id`)

    );

    --成绩表

    CREATE TABLE `Score`(

      `s_id` VARCHAR(20),

      `c_id`  VARCHAR(20),

      `s_score` INT(3),

      PRIMARY KEY(`s_id`,`c_id`)

    );

    --插入学生表测试数据

    insert into Student values('01' , '赵雷' , '1990-01-01' , '男');

    insert into Student values('02' , '钱电' , '1990-12-21' , '男');

    insert into Student values('03' , '孙风' , '1990-05-20' , '男');

    insert into Student values('04' , '李云' , '1990-08-06' , '男');

    insert into Student values('05' , '周梅' , '1991-12-01' , '女');

    insert into Student values('06' , '吴兰' , '1992-03-01' , '女');

    insert into Student values('07' , '郑竹' , '1989-07-01' , '女');

    insert into Student values('08' , '王菊' , '1990-01-20' , '女');

    --课程表测试数据

    insert into Course values('01' , '语文' , '02');

    insert into Course values('02' , '数学' , '01');

    insert into Course values('03' , '英语' , '03');

    --教师表测试数据

    insert into Teacher values('01' , '张三');

    insert into Teacher values('02' , '李四');

    insert into Teacher values('03' , '王五');

    --成绩表测试数据

    insert into Score values('01' , '01' , 80);

    insert into Score values('01' , '02' , 90);

    insert into Score values('01' , '03' , 99);

    insert into Score values('02' , '01' , 70);

    insert into Score values('02' , '02' , 60);

    insert into Score values('02' , '03' , 80);

    insert into Score values('03' , '01' , 80);

    insert into Score values('03' , '02' , 80);

    insert into Score values('03' , '03' , 80);

    insert into Score values('04' , '01' , 50);

    insert into Score values('04' , '02' , 30);

    insert into Score values('04' , '03' , 20);

    insert into Score values('05' , '01' , 76);

    insert into Score values('05' , '02' , 87);

    insert into Score values('06' , '01' , 31);

    insert into Score values('06' , '03' , 34);

    insert into Score values('07' , '02' , 89);

    insert into Score values('07' , '03' , 98);

    练习题和sql语句

    -- 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 a.* ,b.s_score as 01_score,c.s_score as 02_score from

    student a

      join score b on a.s_id=b.s_id and b.c_id='01'

      left join score c on a.s_id=c.s_id and c.c_id='02' or c.c_id = NULL where b.s_score>c.s_score

    --也可以这样写

      select a.*,b.s_score as 01_score,c.s_score as 02_score from student      a,score b,score c

          where a.s_id=b.s_id

          and a.s_id=c.s_id

          and b.c_id='01'

          and c.c_id='02'

          and b.s_score>c.s_score

    -- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数 

    select a.* ,b.s_score as 01_score,c.s_score as 02_score from

      student a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL

      join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.s_score

    -- 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 b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from

      student b

      join score a on b.s_id = a.s_id

      GROUP BY b.s_id,b.s_name HAVING avg_score >=60;

    -- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩

        -- (包括有成绩的和无成绩的)   

    select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from

      student b

      left join score a on b.s_id = a.s_id

      GROUP BY b.s_id,b.s_name HAVING avg_score <60

      union

    select a.s_id,a.s_name,0 as avg_score from

      student a

      where a.s_id not in (

            select distinct s_id from score);

    -- 5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩

    select a.s_id,a.s_name,count(b.c_id) as sum_course,sum(b.s_score) as sum_score from

      student a

      left join score b on a.s_id=b.s_id

      GROUP BY a.s_id,a.s_name;     

    -- 6、查询"李"姓老师的数量

    select count(t_id) from teacher where t_name like '李%';

    -- 7、查询学过"张三"老师授课的同学的信息

    select a.* from

      student a

      join score b on a.s_id=b.s_id where b.c_id in(

        select c_id from course where t_id =(

          select t_id from teacher where t_name = '张三'));

    -- 8、查询没学过"张三"老师授课的同学的信息

    select * from

        student c

        where c.s_id not in(

            select a.s_id from student a join score b on a.s_id=b.s_id where b.c_id in(

            select a.c_id from course a join teacher b on a.t_id = b.t_id where t_name ='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

    select a.* from

      student a,score b,score c

      where a.s_id = b.s_id  and a.s_id = c.s_id and b.c_id='01' and c.c_id='02';

    -- 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 a.* from

      student a

      where a.s_id in (select s_id from score where c_id='01' ) and a.s_id not in(select s_id from score where c_id='02')

    -- 11、查询没有学全所有课程的同学的信息

    --@wendiepei的写法

    select s.* from student s

    left join Score s1 on s1.s_id=s.s_id

    group by s.s_id having count(s1.c_id)<(select count(*) from course) 

    --@k1051785839的写法

    select *

    from student

    where s_id not in(

    select s_id from score t1 

    group by s_id having count(*) =(select count(distinct c_id)  from course))

    -- 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 * from student where s_id in(

      select distinct a.s_id from score a where a.c_id in(select a.c_id from score a where a.s_id='01')

      );

    -- 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

    --@ouyang_1993的写法

    SELECT

    Student.*

    FROM

    Student

    WHERE

    s_id IN (SELECT s_id FROM Score GROUP BY s_id HAVING COUNT(s_id) = (

        #下面的语句是找到'01'同学学习的课程数

        SELECT COUNT(c_id) FROM Score WHERE s_id = '01'

      )

    )

    AND s_id NOT IN (

    #下面的语句是找到学过‘01’同学没学过的课程,有哪些同学。并排除他们

    SELECT s_id FROM Score

    WHERE c_id IN(

      #下面的语句是找到‘01’同学没学过的课程

      SELECT DISTINCT c_id FROM Score

      WHERE c_id NOT IN (

        #下面的语句是找出‘01’同学学习的课程

        SELECT c_id FROM Score WHERE s_id = '01'

    )

      ) GROUP BY s_id

    ) #下面的条件是排除01同学

    AND s_id NOT IN ('01')

    --@k1051785839的写法

    SELECT

    t3.*

    FROM

    (

      SELECT

      s_id,

      group_concat(c_id ORDER BY c_id) group1

      FROM

      score

      WHERE

      s_id &lt;> '01'

      GROUP BY

      s_id

    ) t1

    INNER JOIN (

    SELECT

      group_concat(c_id ORDER BY c_id) group2

    FROM

      score

    WHERE

      s_id = '01'

    GROUP BY

      s_id

    ) t2 ON t1.group1 = t2.group2

    INNER JOIN student t3 ON t1.s_id = t3.s_id

    -- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名

    select a.s_name from student a where a.s_id not in (

      select s_id from score where c_id =

            (select c_id from course where t_id =(

              select t_id from teacher where t_name = '张三')));

    -- 15、查询两门及其以上不及格课程的同学的学号,姓名及其平均成绩

    select a.s_id,a.s_name,ROUND(AVG(b.s_score)) from

      student a

      left join score b on a.s_id = b.s_id

      where a.s_id in(

          select s_id from score where s_score<60 GROUP BY  s_id having count(1)>=2)

      GROUP BY a.s_id,a.s_name

    -- 16、检索"01"课程分数小于60,按分数降序排列的学生信息

    select a.*,b.c_id,b.s_score from

      student a,score b

      where a.s_id = b.s_id and b.c_id='01' and b.s_score<60 ORDER BY b.s_score DESC;

    -- 17、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩

    select a.s_id,(select s_score from score where s_id=a.s_id and c_id='01') as 语文,

            (select s_score from score where s_id=a.s_id and c_id='02') as 数学,

            (select s_score from score where s_id=a.s_id and c_id='03') as 英语,

          round(avg(s_score),2) as 平均分 from score a  GROUP BY a.s_id ORDER BY 平均分 DESC;

    --@喝完这杯还有一箱的写法

    SELECT a.s_id,MAX(CASE a.c_id WHEN '01' THEN a.s_score END ) 语文,

    MAX(CASE a.c_id WHEN '02' THEN a.s_score END ) 数学,

    MAX(CASE a.c_id WHEN '03' THEN a.s_score END ) 英语,

    avg(a.s_score),b.s_name FROM Score a JOIN Student b ON a.s_id=b.s_id GROUP BY a.s_id ORDER BY 5 DESC   

    -- 18.查询各科成绩最高分、最低分和平均分:以如下形式显示:课程ID,课程name,最高分,最低分,平均分,及格率,中等率,优良率,优秀率

    --及格为>=60,中等为:70-80,优良为:80-90,优秀为:>=90

    select a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2),

      ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率,

      ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率,

      ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率,

      ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率

      from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name

    -- 19、按各科成绩进行排序,并显示排名

    -- mysql没有rank函数

      select a.s_id,a.c_id,

            @i:=@i +1 as i保留排名,

            @k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名,

            @score:=a.s_score as score

        from (

            select s_id,c_id,s_score from score GROUP BY s_id,c_id,s_score ORDER BY s_score DESC

    )a,(select @k:=0,@i:=0,@score:=0)s

    --@k1051785839的写法

    (select * from (select

    t1.c_id,

    t1.s_score,

    (select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='01') rank

    FROM score t1 where t1.c_id='01'

    order by t1.s_score desc) t1)

    union

    (select * from (select

    t1.c_id,

    t1.s_score,

    (select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='02') rank

    FROM score t1 where t1.c_id='02'

    order by t1.s_score desc) t2)

    union

    (select * from (select

    t1.c_id,

    t1.s_score,

    (select count(distinct t2.s_score) from score t2 where t2.s_score>=t1.s_score and t2.c_id='03') rank

    FROM score t1 where t1.c_id='03'

    order by t1.s_score desc) t3)

    -- 20、查询学生的总成绩并进行排名

    select a.s_id,

      @i:=@i+1 as i,

      @k:=(case when @score=a.sum_score then @k else @i end) as rank,

      @score:=a.sum_score as score

    from (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a,

      (select @k:=0,@i:=0,@score:=0)s

    -- 21、查询不同老师所教不同课程平均分从高到低显示

      select a.t_id,c.t_name,a.c_id,ROUND(avg(s_score),2) as avg_score from course a

        left join score b on a.c_id=b.c_id

        left join teacher c on a.t_id=c.t_id

        GROUP BY a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;

    -- 22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩

          select d.*,c.排名,c.s_score,c.c_id from (

                    select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01' 

                    ORDER BY a.s_score DESC 

                )c

                left join student d on c.s_id=d.s_id

                where 排名 BETWEEN 2 AND 3

                UNION

                select d.*,c.排名,c.s_score,c.c_id from (

                    select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02' 

                    ORDER BY a.s_score DESC

                )c

                left join student d on c.s_id=d.s_id

                where 排名 BETWEEN 2 AND 3

                UNION

                select d.*,c.排名,c.s_score,c.c_id from (

                    select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03'

                    ORDER BY a.s_score DESC

                )c

                left join student d on c.s_id=d.s_id

                where 排名 BETWEEN 2 AND 3;

    -- 23、统计各科成绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比

        select distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score a

            left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`,

                          ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比

                    from score GROUP BY c_id)b on a.c_id=b.c_id

            left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`,

                          ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比

                    from score GROUP BY c_id)c on a.c_id=c.c_id

            left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`,

                          ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比

                    from score GROUP BY c_id)d on a.c_id=d.c_id

            left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`,

                          ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比

                    from score GROUP BY c_id)e on a.c_id=e.c_id

            left join course f on a.c_id = f.c_id

    -- 24、查询学生平均成绩及其名次

        select a.s_id,

            @i:=@i+1 as '不保留空缺排名',

            @k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名',

            @avg_score:=avg_s as '平均分'

        from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id ORDER BY avg_s DESC)a,(select @avg_score:=0,@i:=0,@k:=0)b;

    -- 25、查询各科成绩前三名的记录

          -- 1.选出b表比a表成绩大的所有组

          -- 2.选出比当前id成绩大的 小于三个的

        select a.s_id,a.c_id,a.s_score from score a

          left join score b on a.c_id = b.c_id and a.s_score<b.s_score

          group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3

          ORDER BY a.c_id,a.s_score DESC

    -- 26、查询每门课程被选修的学生数

        select c_id,count(s_id) from score a GROUP BY c_id

    -- 27、查询出只有两门课程的全部学生的学号和姓名

        select s_id,s_name from student where s_id in(

            select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2);

    -- 28、查询男生、女生人数

        select s_sex,COUNT(s_sex) as 人数  from student GROUP BY s_sex

    -- 29、查询名字中含有"风"字的学生信息

        select * from student where s_name like '%风%';

    -- 30、查询同名同性学生名单,并统计同名人数   

        select a.s_name,a.s_sex,count(*) from student a  JOIN

              student b on a.s_id !=b.s_id and a.s_name = b.s_name and a.s_sex = b.s_sex

        GROUP BY a.s_name,a.s_sex

    -- 31、查询1990年出生的学生名单   

        select s_name from student where s_birth like '1990%'

    -- 32、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 c_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC

    -- 33、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 a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a

        left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85

    -- 34、查询课程名称为"数学",且分数低于60的学生姓名和分数 

        select a.s_name,b.s_score from score b join student a on a.s_id=b.s_id where b.c_id=(

              select c_id from course where c_name ='数学') and b.s_score<60

    -- 35、查询所有学生的课程及分数情况;

        select a.s_id,a.s_name,

              SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文',

              SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学',

              SUM(case c.c_name when '英语' then b.s_score else 0 end) as '英语',

              SUM(b.s_score) as  '总分'

        from student a left join score b on a.s_id = b.s_id

        left join course c on b.c_id = c.c_id

        GROUP BY a.s_id,a.s_name

    -- 36、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 a.s_name,b.c_name,c.s_score from course b left join score c on b.c_id = c.c_id

            left join student a on a.s_id=c.s_id where c.s_score>=70

    -- 37、查询不及格的课程

        select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_id

          where a.s_score<60

    -- 38、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 a.s_id,b.s_name from score a LEFT JOIN student b on a.s_id = b.s_id

          where a.c_id = '01'  and a.s_score>80

    -- 39、求每门课程的学生人数

        select count(*) from score GROUP BY c_id;

    -- 40、查询选修"张三"老师所授课程的学生中,成绩最高的学生信息及其成绩

        -- 查询老师id 

        select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三'

        -- 查询最高分(可能有相同分数)

        select MAX(s_score) from score where c_id='02'

        -- 查询信息

        select a.*,b.s_score,b.c_id,c.c_name from student a

          LEFT JOIN score b on a.s_id = b.s_id

          LEFT JOIN course c on b.c_id=c.c_id

          where b.c_id =(select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三')

          and b.s_score in (select MAX(s_score) from score where c_id='02')

    -- 41、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 DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score

    -- 42、查询每门功成绩最好的前两名

        -- 牛逼的写法

      select a.s_id,a.c_id,a.s_score from score a

        where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id

    -- 43、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 c_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC

    -- 44、检索至少选修两门课程的学生学号

        select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2

    -- 45、查询选修了全部课程的学生信息

        select * from student where s_id in(   

          select s_id from score GROUP BY s_id HAVING count(*)=(select count(*) from course))

    -- 46、查询各学生的年龄

      -- 按照出生日期来算,当前月日 < 出生年月的月日则,年龄减一

      select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') -

            (case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as age

        from student;

    -- 47、查询本周过生日的学生

      select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)

      select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

      select WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))

    -- 48、查询下周过生日的学生

      select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth)

    -- 49、查询本月过生日的学生

      select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)

    -- 50、查询下月过生日的学生

    select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)

    原文:https://mp.weixin.qq.com/s/xIU3jRiF8KTt8JxR3vlQtw

    作者:Java资源库

    来源:微信公众号

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