MySQL练习题及答案

一 题目

1、查询所有的课程的名称以及对应的任课老师姓名

2、查询学生表中男女生各有多少人

3、查询物理成绩等于100的学生的姓名

4、查询平均成绩大于八十分的同学的姓名和平均成绩

5、查询所有学生的学号,姓名,选课数,总成绩

6、 查询姓李老师的个数

7、 查询没有报李平老师课的学生姓名

8、 查询物理课程比生物课程高的学生的学号

9、 查询没有同时选修物理课程和体育课程的学生姓名

10、查询挂科超过两门(包括两门)的学生姓名和班级
、查询选修了所有课程的学生姓名

12、查询李平老师教的课程的所有成绩记录
 
13、查询全部学生都选修了的课程号和课程名

14、查询每门课程被选修的次数

15、查询之选修了一门课程的学生姓名和学号

16、查询所有学生考出的成绩并按从高到低排序(成绩去重)

17、查询平均成绩大于85的学生姓名和平均成绩

18、查询生物成绩不及格的学生姓名和对应生物分数

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

20、查询每门课程成绩最好的前两名学生姓名

21、查询不同课程但成绩相同的学号,课程号,成绩

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;

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;

24、任课最多的老师中学生单科成绩最高的学生姓名

二 答案

#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 性别,
    count(1) 人数
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(注意:对于那些没有选修任何课程的学生也算在内)
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(找出报名李平老师课程的学生,然后取反就可以)
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(分别得到物理成绩表与生物成绩表,然后连表即可)
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;




#9、 查询没有同时选修物理课程和体育课程的学生姓名(没有同时选修指的是选修了一门的,思路是得到物理+体育课程的学生信息表,然后基于学生分组,统计count(课程)=1)
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(求出<60的表,然后对学生进行分组,统计课程数目>=2)
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(先从course表统计课程的总数,然后基于score表按照student_id分组,统计课程数据等于课程总数即可)
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
SELECT
    *
FROM
    score
WHERE
    course_id IN (
        SELECT
            cid
        FROM
            course
        INNER JOIN teacher ON course.teacher_id = teacher.tid
        WHERE
            teacher.tname = '李平老师'
    );




#13、查询全部学生都选修了的课程号和课程名(取所有学生数,然后基于score表的课程分组,找出count(student_id)等于学生数即可)
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
            )
    );




#14、查询每门课程被选修的次数
SELECT
    course_id,
    COUNT(student_id)
FROM
    score
GROUP BY
    course_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
    num
FROM
    score
ORDER BY
    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
#查看每门课程按照分数排序的信息,为下列查找正确与否提供依据
SELECT
    *
FROM
    score
ORDER BY
    course_id,
    num DESC;




#表1:求出每门课程的课程course_id,与最高分数first_num
SELECT
    course_id,
    max(num) first_num
FROM
    score
GROUP BY
    course_id;




#表2:去掉最高分,再按照课程分组,取得的最高分,就是第二高的分数second_num
SELECT
    score.course_id,
    max(num) second_num
FROM
    score
INNER JOIN (
    SELECT
        course_id,
        max(num) first_num
    FROM
        score
    GROUP BY
        course_id
) AS t ON score.course_id = t.course_id
WHERE
    score.num < t.first_num
GROUP BY
    course_id;




#将表1和表2联合到一起,得到一张表t3,包含课程course_id与该们课程的first_num与second_num
SELECT
    t1.course_id,
    t1.first_num,
    t2.second_num
FROM
    (
        SELECT
            course_id,
            max(num) first_num
        FROM
            score
        GROUP BY
            course_id
    ) AS t1
INNER JOIN (
    SELECT
        score.course_id,
        max(num) second_num
    FROM
        score
    INNER JOIN (
        SELECT
            course_id,
            max(num) first_num
        FROM
            score
        GROUP BY
            course_id
    ) AS t ON score.course_id = t.course_id
    WHERE
        score.num < t.first_num
    GROUP BY
        course_id
) AS t2 ON t1.course_id = t2.course_id;




#查询前两名的学生(有可能出现并列第一或者并列第二的情况)
SELECT
    score.student_id,
    t3.course_id,
    t3.first_num,
    t3.second_num
FROM
    score
INNER JOIN (
    SELECT
        t1.course_id,
        t1.first_num,
        t2.second_num
    FROM
        (
            SELECT
                course_id,
                max(num) first_num
            FROM
                score
            GROUP BY
                course_id
        ) AS t1
    INNER JOIN (
        SELECT
            score.course_id,
            max(num) second_num
        FROM
            score
        INNER JOIN (
            SELECT
                course_id,
                max(num) first_num
            FROM
                score
            GROUP BY
                course_id
        ) AS t ON score.course_id = t.course_id
        WHERE
            score.num < t.first_num
        GROUP BY
            course_id
    ) AS t2 ON t1.course_id = t2.course_id
) AS t3 ON score.course_id = t3.course_id
WHERE
    score.num >= t3.second_num
AND score.num <= t3.first_num;




#排序后可以看的明显点
SELECT
    score.student_id,
    t3.course_id,
    t3.first_num,
    t3.second_num
FROM
    score
INNER JOIN (
    SELECT
        t1.course_id,
        t1.first_num,
        t2.second_num
    FROM
        (
            SELECT
                course_id,
                max(num) first_num
            FROM
                score
            GROUP BY
                course_id
        ) AS t1
    INNER JOIN (
        SELECT
            score.course_id,
            max(num) second_num
        FROM
            score
        INNER JOIN (
            SELECT
                course_id,
                max(num) first_num
            FROM
                score
            GROUP BY
                course_id
        ) AS t ON score.course_id = t.course_id
        WHERE
            score.num < t.first_num
        GROUP BY
            course_id
    ) AS t2 ON t1.course_id = t2.course_id
) AS t3 ON score.course_id = t3.course_id
WHERE
    score.num >= t3.second_num
AND score.num <= t3.first_num
ORDER BY
    course_id;




#可以用以下命令验证上述查询的正确性
SELECT
    *
FROM
    score
ORDER BY
    course_id,
    num DESC;




-- 21、查询不同课程但成绩相同的学号,课程号,成绩
-- 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;
-- 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;
-- 24、任课最多的老师中学生单科成绩最高的学生姓名

更多练习以及参考答案:https://www.cnblogs.com/clschao/articles/9995768.html

posted @ 2019-09-21 09:58  changxin7  阅读(...)  评论(...编辑  收藏