mysql题目练习的答案

rianley 博客              盗版必究！
基于上一个sql练习的答案；该答案是我所写！如果有错误或者用法不当之处；请下方评论指出；感激不尽！谢谢各位码友！


<!--1、查询所有的课程的名称以及对应的任课老师姓名 -->

SELECT
    course.cname,
    teacher.tname
FROM
    course
INNER JOIN teacher ON course.teacher_id = teacher.tid;

　　

<!--2、查询学生表中男女生各有多少人-->

SELECT
    gender
    count(1)
FROM
    student
GROUP BY
    gender;

　　

<!--3、查询物理成绩等于100的学生的姓名-->

SELECT
    student.sname
FROM
    student
WHERE
    sid IN (
        SELECT
            student_id
        FROM
            score
        INNER JOIN course ON score.course_id = course.cid
        WHERE
            course.cname = '物理'
        AND score.num = 100
    );

　　

<!--4、查询平均成绩大于八十分的同学的姓名和平均成绩-->

SELECT
    student.sname,
    t1.avg_num
FROM
    student
INNER JOIN (
    SELECT
        student_id,
        avg(num) AS avg_num
    FROM
        score
    GROUP BY
        student_id
    HAVING
        avg(num) > 80
) AS t1 ON student.sid = t1.student_id;

　　

<!--5、查询所有学生的学号，姓名，选课数，总成绩 -->

SELECT
    student.sid,
    student.sname,
    t1.course_num,
    t1.total_num
FROM
    student
LEFT JOIN (
    SELECT
        student_id,
        COUNT(course_id) course_num,
        sum(num) total_num
    FROM
        score
    GROUP BY
        student_id
) AS t1 ON student.sid = t1.student_id;

　　

<!--6、 查询姓李老师的个数-->

SELECT
    count(tid)
FROM
    teacher
WHERE
    tname LIKE '李%';

　　

<!--7、 查询没有报李平老师课的学生姓名 -->

SELECT
    student.sname
FROM
    student
WHERE
    sid NOT IN (
        SELECT DISTINCT
            student_id
        FROM
            score
        WHERE
            course_id IN (
                SELECT
                    course.cid
                FROM
                    course
                INNER JOIN teacher ON course.teacher_id = teacher.tid
                WHERE
                    teacher.tname = '李平老师'
            )
    );

　　

<!--8、 查询物理课程比生物课程高的学生的学号 -->

SELECT
    t1.student_id
FROM
    (
        SELECT
            student_id,
            num
        FROM
            score
        WHERE
            course_id = (
                SELECT
                    cid
                FROM
                    course
                WHERE
                    cname = '物理'
            )
    ) AS t1
INNER JOIN (
    SELECT
        student_id,
        num
    FROM
        score
    WHERE
        course_id = (
            SELECT
                cid
            FROM
                course
            WHERE
                cname = '生物'
        )
) AS t2 ON t1.student_id = t2.student_id
WHERE
    t1.num > t2.num;

　　

<!--方法二-->

SELECT x.sname from
        (SELECT c.sid,c.sname,b.num from course as a INNER JOIN score as b on a.cid=b.course_id
        and a.cname ='物理'
        LEFT JOIN student as c on b.student_id=c.sid)
        as x INNER JOIN
        (SELECT c.sid,c.sname,b.num from course as a INNER JOIN score as b on a.cid=b.course_id
        and a.cname ='生物'
        LEFT JOIN student as c on b.student_id=c.sid)
        as y on x.sid=y.sid WHERE x.num>y.num

　　

<!--9、 查询没有同时选修物理课程和体育课程的学生姓名 -->

SELECT
    student.sname
FROM
    student
WHERE
    sid IN (
        SELECT
            student_id
        FROM
            score
        WHERE
            course_id IN (
                SELECT
                    cid
                FROM
                    course
                WHERE
                    cname = '物理'
                OR cname = '体育'
            )
        GROUP BY
            student_id
        HAVING
            COUNT(course_id) = 1
    );

　　

<!--10、查询挂科超过两门(包括两门)的学生姓名和班级 -->

SELECT
    student.sname,
    class.caption
FROM
    student
INNER JOIN (
    SELECT
        student_id
    FROM
        score
    WHERE
        num < 60
    GROUP BY
        student_id
    HAVING
        count(course_id) >= 2
) AS t1
INNER JOIN class ON student.sid = t1.student_id
AND student.class_id = class.cid;

　　

<!--11、查询选修了所有课程的学生姓名 -->

SELECT
    student.sname
FROM
    student
WHERE
    sid IN (
        SELECT
            student_id
        FROM
            score
        GROUP BY
            student_id
        HAVING
            COUNT(course_id) = (SELECT count(cid) FROM course)
    );

　　

<!--12、查询李平老师教的课程的所有成绩记录-->
    <!-- 方法一 -->

select
        score.num
     from
        score
     join
        (select cid from course inner join teacher on course.teacher_id = teacher.tid where tid =2) as t1
      on
          score.course_id=t1.cid;

　　

    <!-- 方法二 -->

 select
      score.num
    from
        score
    where
        course_id
            in (select cid from course inner join teacher on course.teacher_id = teacher.tid where tid =2)

　　

<!--13、查询全部学生都选修了的课程号和课程名-->

<!-- 方式一 -->

 SELECT
        cid,
        cname
    FROM
        course
    WHERE
        cid IN ( select course_id from score group by course_id having COUNT(student_id) = (select count(sid) from student)  )  <!-- 注解：--先将所有的课程分组 然后 过滤学生没有全选择的课程，在通过cid 找出 对应的课程号与课程名-->
        );

　　

<!--方式二-->

 SELECT course.cname,count(course.cid) as count_id from course INNER JOIN score  on course.cid=score.course_id
              group by course.cid HAVING count_id=(SELECT count(sid) from student);

　　

<!--方式三-->

  SELECT a.cid,a.cname from course as a INNER JOIN score as b on a.cid=b.course_id
            GROUP BY  a.cid HAVING count(a.cid)=(SELECT count(sid) from student);

　　

<!--14、查询每门课程被选修的次数-->

select course_id,count(student_id) from score group by student_id;

　　

<!--15、查询之选修了一门课程的学生姓名和学号-->

SELECT
    sid,
    sname
FROM
    student
WHERE
    sid IN (
        SELECT
            student_id
        FROM
            score
        GROUP BY
            student_id
        HAVING
            COUNT(course_id) = 1
    );

　　

<!--16、查询所有学生考出的成绩并按从高到低排序（成绩去重）-->

SELECT DISTINCT sum(if(b.num is null,0,b.num)) as all_num from
        student as a LEFT JOIN score as b on a.sid=b.student_id
        GROUP BY a.sid  ORDER BY all_num DESC

　　

<!--17、查询平均成绩大于85的学生姓名和平均成绩-->

SELECT
    sname,
    t1.avg_num
FROM
    student
INNER JOIN (
    SELECT
        student_id,
        avg(num) avg_num
    FROM
        score
    GROUP BY
        student_id
    HAVING
        AVG(num) > 85
) t1 ON student.sid = t1.student_id;

　　

<!--18、查询生物成绩不及格的学生姓名和对应生物分数-->

SELECT
    sname 姓名,
    num 生物成绩
FROM
    score
LEFT JOIN course ON score.course_id = course.cid
LEFT JOIN student ON score.student_id = student.sid
WHERE
    course.cname = '生物'
AND score.num < 60;

　　

<!--19、查询在所有选修了李平老师课程的学生中，这些课程(李平老师的课程，不是所有课程)平均成绩最高的学生姓名-->

SELECT
    sname
FROM
    student
WHERE
    sid = (
        SELECT
            student_id
        FROM
            score
        WHERE
            course_id IN (
                SELECT
                    course.cid
                FROM
                    course
                INNER JOIN teacher ON course.teacher_id = teacher.tid
                WHERE
                    teacher.tname = '李平老师'
            )
        GROUP BY
            student_id
        ORDER BY
            AVG(num) DESC
        LIMIT 1
    );

　　

<!--20、查询每门课程成绩最好的前两名学生姓名-->

SELECT c.course_id,c.student_id,c.num
        from
        score as c LEFT JOIN
        (SELECT b.course_id,max(num) as second_num from (SELECT course_id,max(num) first_num from score GROUP BY course_id) as a LEFT JOIN score as b
        on a.course_id=b.course_id WHERE a.first_num>b.num GROUP BY b.course_id) as b on c.course_id=b.course_id and b.second_num=c.num

        LEFT JOIN
        (SELECT b.course_id,b.student_id,a.first_num from (SELECT course_id,max(num) first_num from score GROUP BY course_id) as a LEFT JOIN score as b on a.course_id=b.course_id and a.first_num=b.num) as a
        on c.course_id=a.course_id and a.first_num=c.num

        WHERE a.first_num is NOT NULL or b.second_num is NOT NULL
        GROUP BY c.student_id,c.num
        ORDER BY c.course_id,c.num desc

　　

<!--21、查询不同课程但成绩相同的学号，课程号，成绩-->

SELECT a.student_id,a.course_id,a.num from (SELECT num,course_id,student_id FROM score
        GROUP BY course_id,num)  as a
        GROUP BY a.num
        HAVING count(a.num)>1

　　

<!--22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称；-->

SELECT DISTINCT d.sname,b.cname from teacher as a INNER JOIN course as b
        on a.tid=b.teacher_id and a.tname!='叶平老师'
        LEFT JOIN score as c on c.course_id=b.cid
        LEFT JOIN student as d on d.sid=c.student_id

　　

<!--23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名；-->

SELECT DISTINCT b.sid,b.sname FROM score as a  LEFT JOIN student as b
        on a.student_id=b.sid
        WHERE a.course_id in (SELECT course_id FROM score WHERE student_id=1)

　　

<!--24、任课最多的老师中学生单科成绩最高的学生姓名-->

SELECT b.cname,b.num,c.sname from    <!-- 注：这里没有考虑2个老师任课数目一样，考虑了单科成绩可能会一样-->
        (SELECT a.cname,max(a.num) as max_num from (
        SELECT a.cname,b.num,b.student_id from course as a LEFT JOIN score as b on a.cid=b.course_id
        WHERE a.teacher_id =(SELECT COUNT(teacher_id) as teacher_idx  FROM course GROUP BY teacher_id
        ORDER BY  teacher_idx desc LIMIT 1)
        ) as a
        GROUP BY a.cname) as a LEFT JOIN (
        SELECT a.cname,b.num,b.student_id from course as a LEFT JOIN score as b on a.cid=b.course_id
        WHERE a.teacher_id =(SELECT COUNT(teacher_id) as teacher_idx  FROM course GROUP BY teacher_id
        ORDER BY  teacher_idx desc LIMIT 1)
        ) as b on a.cname=b.cname and a.max_num=b.num
        LEFT JOIN student as c on c.sid=b.student_id

　　

<!------------------------------------------------------------->
<!--扩展题-->
<!--  问题来了，有个上班打卡表，记录公司多个user的打开记录（每天至少打卡2次），表里字段：id,user_id,sign_time，要求统计出2018-6月份user迟到情况,9点以后算迟到  -->

CREATE TABLE  sign_in(id INT NOT NULL  PRIMARY KEY AUTO_INCREMENT,
               `user_id` int not  null ,
                sign_time datetime   not  null )ENGINE=INNODB CHARSET=UTF8

　　

<!--迟到情况 大于9点算是迟到--用substr 获取>

select a.sign_time,a.user_id from sign_in as a inner join (select id,sign_time from sign_in  where   sign_time >='2018-6-1 00:00:00' and  sign_time < '2018-7-1 00:00:00'
          GROUP BY user_id,DATE(sign_time) having  substr(sign_time,12,2) > '09') as t1
          on a.id= t1.id

　　

<!--迟到情况 大于9点算是迟到-- 用 time 获取>

SELECT user_id, MIN(sign_time) as min_sign_time from sign_in WHERE sign_time like '2018-06%' GROUP BY user_id,DATE(sign_time)  HAVING TIME(min_sign_time)>'09:00:00'

　　

<!--统计出上个月上班不足8小时的user的情况-->

SELECT user_id, MIN(sign_time) as min_sign_time,MAX(sign_time) as max_sign_time  from sign_in WHERE sign_time like '2018-06%' GROUP BY user_id,DATE(sign_time)
HAVING UNIX_TIMESTAMP(max_sign_time)-UNIX_TIMESTAMP(min_sign_time)<8*3600

　　

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