mysql 练习题答案

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

MySQL练习题参考答案

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导入现有数据库数据：

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/*
 Navicat Premium Data Transfer

 Source Server         : localhost
 Source Server Type    : MySQL
 Source Server Version : 50624
 Source Host           : localhost
 Source Database       : sqlexam

 Target Server Type    : MySQL
 Target Server Version : 50624
 File Encoding         : utf-8

 Date: 10/21/2016 06:46:46 AM
*/

SET NAMES utf8;
SET FOREIGN_KEY_CHECKS = 0;

-- ----------------------------
--  Table structure for `class`
-- ----------------------------
DROP TABLE IF EXISTS `class`;
CREATE TABLE `class` (
  `cid` int(11) NOT NULL AUTO_INCREMENT,
  `caption` varchar(32) NOT NULL,
  PRIMARY KEY (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `class`
-- ----------------------------
BEGIN;
INSERT INTO `class` VALUES ('1', '三年二班'), ('2', '三年三班'), ('3', '一年二班'), ('4', '二年九班');
COMMIT;

-- ----------------------------
--  Table structure for `course`
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
  `cid` int(11) NOT NULL AUTO_INCREMENT,
  `cname` varchar(32) NOT NULL,
  `teacher_id` int(11) NOT NULL,
  PRIMARY KEY (`cid`),
  KEY `fk_course_teacher` (`teacher_id`),
  CONSTRAINT `fk_course_teacher` FOREIGN KEY (`teacher_id`) REFERENCES `teacher` (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `course`
-- ----------------------------
BEGIN;
INSERT INTO `course` VALUES ('1', '生物', '1'), ('2', '物理', '2'), ('3', '体育', '3'), ('4', '美术', '2');
COMMIT;

-- ----------------------------
--  Table structure for `score`
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score` (
  `sid` int(11) NOT NULL AUTO_INCREMENT,
  `student_id` int(11) NOT NULL,
  `course_id` int(11) NOT NULL,
  `num` int(11) NOT NULL,
  PRIMARY KEY (`sid`),
  KEY `fk_score_student` (`student_id`),
  KEY `fk_score_course` (`course_id`),
  CONSTRAINT `fk_score_course` FOREIGN KEY (`course_id`) REFERENCES `course` (`cid`),
  CONSTRAINT `fk_score_student` FOREIGN KEY (`student_id`) REFERENCES `student` (`sid`)
) ENGINE=InnoDB AUTO_INCREMENT=53 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `score`
-- ----------------------------
BEGIN;
INSERT INTO `score` VALUES ('1', '1', '1', '10'), ('2', '1', '2', '9'), ('5', '1', '4', '66'), ('6', '2', '1', '8'), ('8', '2', '3', '68'), ('9', '2', '4', '99'), ('10', '3', '1', '77'), ('11', '3', '2', '66'), ('12', '3', '3', '87'), ('13', '3', '4', '99'), ('14', '4', '1', '79'), ('15', '4', '2', '11'), ('16', '4', '3', '67'), ('17', '4', '4', '100'), ('18', '5', '1', '79'), ('19', '5', '2', '11'), ('20', '5', '3', '67'), ('21', '5', '4', '100'), ('22', '6', '1', '9'), ('23', '6', '2', '100'), ('24', '6', '3', '67'), ('25', '6', '4', '100'), ('26', '7', '1', '9'), ('27', '7', '2', '100'), ('28', '7', '3', '67'), ('29', '7', '4', '88'), ('30', '8', '1', '9'), ('31', '8', '2', '100'), ('32', '8', '3', '67'), ('33', '8', '4', '88'), ('34', '9', '1', '91'), ('35', '9', '2', '88'), ('36', '9', '3', '67'), ('37', '9', '4', '22'), ('38', '10', '1', '90'), ('39', '10', '2', '77'), ('40', '10', '3', '43'), ('41', '10', '4', '87'), ('42', '11', '1', '90'), ('43', '11', '2', '77'), ('44', '11', '3', '43'), ('45', '11', '4', '87'), ('46', '12', '1', '90'), ('47', '12', '2', '77'), ('48', '12', '3', '43'), ('49', '12', '4', '87'), ('52', '13', '3', '87');
COMMIT;

-- ----------------------------
--  Table structure for `student`
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (
  `sid` int(11) NOT NULL AUTO_INCREMENT,
  `gender` char(1) NOT NULL,
  `class_id` int(11) NOT NULL,
  `sname` varchar(32) NOT NULL,
  PRIMARY KEY (`sid`),
  KEY `fk_class` (`class_id`),
  CONSTRAINT `fk_class` FOREIGN KEY (`class_id`) REFERENCES `class` (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=17 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `student`
-- ----------------------------
BEGIN;
INSERT INTO `student` VALUES ('1', '男', '1', '理解'), ('2', '女', '1', '钢蛋'), ('3', '男', '1', '张三'), ('4', '男', '1', '张一'), ('5', '女', '1', '张二'), ('6', '男', '1', '张四'), ('7', '女', '2', '铁锤'), ('8', '男', '2', '李三'), ('9', '男', '2', '李一'), ('10', '女', '2', '李二'), ('11', '男', '2', '李四'), ('12', '女', '3', '如花'), ('13', '男', '3', '刘三'), ('14', '男', '3', '刘一'), ('15', '女', '3', '刘二'), ('16', '男', '3', '刘四');
COMMIT;

-- ----------------------------
--  Table structure for `teacher`
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (
  `tid` int(11) NOT NULL AUTO_INCREMENT,
  `tname` varchar(32) NOT NULL,
  PRIMARY KEY (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;

-- ----------------------------
--  Records of `teacher`
-- ----------------------------
BEGIN;
INSERT INTO `teacher` VALUES ('1', '张磊老师'), ('2', '李平老师'), ('3', '刘海燕老师'), ('4', '朱云海老师'), ('5', '李杰老师');
COMMIT;

SET FOREIGN_KEY_CHECKS = 1;

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2、查询“生物”课程比“物理”课程成绩高的所有学生的学号； 思路：     获取所有有生物课程的人（学号，成绩） - 临时表     获取所有有物理课程的人（学号，成绩） - 临时表     根据【学号】连接两个临时表：         学号  物理成绩   生物成绩       然后再进行筛选           select A.student_id,sw,ty from           (select student_id,num as sw from score left join course on score.course_id = course.cid where course.cname = '生物') as A           left join           (select student_id,num  as ty from score left join course on score.course_id = course.cid where course.cname = '体育') as B           on A.student_id = B.student_id where sw > if(isnull(ty),0,ty);   3、查询平均成绩大于60分的同学的学号和平均成绩；     思路：         根据学生分组，使用avg获取平均值，通过having对avg进行筛选           select student_id,avg(num) from score group by student_id having avg(num) > 60   4、查询所有同学的学号、姓名、选课数、总成绩；       select score.student_id,sum(score.num),count(score.student_id),student.sname     from         score left join student on score.student_id = student.sid       group by score.student_id   5、查询姓“李”的老师的个数；     select count(tid) from teacher where tname like '李%'       select count(1) from (select tid from teacher where tname like '李%') as B   6、查询没学过“叶平”老师课的同学的学号、姓名；     思路：         先查到“李平老师”老师教的所有课ID         获取选过课的所有学生ID         学生表中筛选     select * from student where sid not in (         select DISTINCT student_id from score where score.course_id in (             select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '李平老师'         )     )   7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；     思路：         先查到既选择001又选择002课程的所有同学         根据学生进行分组，如果学生数量等于2表示，两门均已选择       select student_id,sname from       (select student_id,course_id from score where course_id = 1 or course_id = 2) as B            left join student on B.student_id = student.sid group by student_id HAVING count(student_id) > 1     8、查询学过“叶平”老师所教的所有课的同学的学号、姓名；       同上，只不过将001和002变成 in (叶平老师的所有课)   9、查询课程编号“002”的成绩比课程编号“001”课程低的所有同学的学号、姓名；     同第1题     10、查询有课程成绩小于60分的同学的学号、姓名；               select sid,sname from student where sid in (         select distinct student_id from score where num < 60     )   11、查询没有学全所有课的同学的学号、姓名；     思路：         在分数表中根据学生进行分组，获取每一个学生选课数量         如果数量 == 总课程数量，表示已经选择了所有课程           select student_id,sname         from score left join student on score.student_id = student.sid         group by student_id HAVING count(course_id) = (select count(1) from course)     12、查询至少有一门课与学号为“001”的同学所学相同的同学的学号和姓名；     思路：         获取 001 同学选择的所有课程         获取课程在其中的所有人以及所有课程         根据学生筛选，获取所有学生信息         再与学生表连接，获取姓名           select student_id,sname, count(course_id)         from score left join student on score.student_id = student.sid         where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id   13、查询至少学过学号为“001”同学所有课的其他同学学号和姓名；         先找到和001的学过的所有人         然后个数 ＝ 001所有学科     ＝＝》 其他人可能选择的更多           select student_id,sname, count(course_id)         from score left join student on score.student_id = student.sid         where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id having count(course_id) ＝ (select count(course_id) from score where student_id = 1)   14、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名；                   个数相同         002学过的也学过           select student_id,sname from score left join student on score.student_id = student.sid where student_id in (             select student_id from score  where student_id != 1 group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)         ) and course_id in (select course_id from score where student_id = 1) group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 1)     15、删除学习“叶平”老师课的score表记录；       delete from score where course_id in (         select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.name = '叶平'     )   16、向SC表中插入一些记录，这些记录要求符合以下条件：①没有上过编号“002”课程的同学学号；②插入“002”号课程的平均成绩；     思路：         由于insert 支持                  inset into tb1(xx,xx) select x1,x2 from tb2;         所有，获取所有没上过002课的所有人，获取002的平均成绩       insert into score(student_id, course_id, num) select sid,2,(select avg(num) from score where course_id = 2)     from student where sid not in (         select student_id from score where course_id = 2     )       17、按平均成绩从低到高 显示所有学生的“语文”、“数学”、“英语”三门的课程成绩，按如下形式显示： 学生ID,语文,数学,英语,有效课程数,有效平均分；     select sc.student_id,         (select num from score left join course on score.course_id = course.cid where course.cname = "生物" and score.student_id=sc.student_id) as sy,         (select num from score left join course on score.course_id = course.cid where course.cname = "物理" and score.student_id=sc.student_id) as wl,         (select num from score left join course on score.course_id = course.cid where course.cname = "体育" and score.student_id=sc.student_id) as ty,         count(sc.course_id),         avg(sc.num)     from score as sc     group by student_id desc           18、查询各科成绩最高和最低的分：以如下形式显示：课程ID，最高分，最低分；           select course_id, max(num) as max_num, min(num) as min_num from score group by course_id;   19、按各科平均成绩从低到高和及格率的百分数从高到低顺序；     思路：case when .. then     select course_id, avg(num) as avgnum,sum(case when score.num > 60 then 1 else 0 END)/count(1)*100 as percent from score group by course_id order by avgnum asc,percent desc;   20、课程平均分从高到低显示（现实任课老师）；       select avg(if(isnull(score.num),0,score.num)),teacher.tname from course     left join score on course.cid = score.course_id     left join teacher on course.teacher_id = teacher.tid       group by score.course_id     21、查询各科成绩前三名的记录:(不考虑成绩并列情况)     select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join     (     select         sid,         (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0,1) as first_num,         (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 3,1) as second_num     from         score as s1     ) as T     on score.sid =T.sid     where score.num <= T.first_num and score.num >= T.second_num   22、查询每门课程被选修的学生数；           select course_id, count(1) from score group by course_id;   23、查询出只选修了一门课程的全部学生的学号和姓名；     select student.sid, student.sname, count(1) from score       left join student on score.student_id  = student.sid        group by course_id having count(1) = 1     24、查询男生、女生的人数；     select * from     (select count(1) as man from student where gender='男') as A ,     (select count(1) as feman from student where gender='女') as B   25、查询姓“张”的学生名单；     select sname from student where sname like '张%';   26、查询同名同姓学生名单，并统计同名人数；       select sname,count(1) as count from student group by sname;   27、查询每门课程的平均成绩，结果按平均成绩升序排列，平均成绩相同时，按课程号降序排列；     select course_id,avg(if(isnull(num), 0 ,num)) as avg from score group by course_id order by avg     asc,course_id desc;   28、查询平均成绩大于85的所有学生的学号、姓名和平均成绩；       select student_id,sname, avg(if(isnull(num), 0 ,num)) from score left join student on score.student_id = student.sid group by student_id;   29、查询课程名称为“数学”，且分数低于60的学生姓名和分数；       select student.sname,score.num from score     left join course on score.course_id = course.cid     left join student on score.student_id = student.sid     where score.num < 60 and course.cname = '生物'   30、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名；     select * from score where score.student_id = 3 and score.num > 80   31、求选了课程的学生人数       select count(distinct student_id) from score       select count(c) from (         select count(student_id) as c from score group by student_id) as A   32、查询选修“杨艳”老师所授课程的学生中，成绩最高的学生姓名及其成绩；           select sname,num from score     left join student on score.student_id = student.sid     where score.course_id in (select course.cid from course left join teacher on course.teacher_id = teacher.tid where tname='张磊老师') order by num desc limit 1;   33、查询各个课程及相应的选修人数；     select course.cname,count(1) from score     left join course on score.course_id = course.cid     group by course_id;     34、查询不同课程但成绩相同的学生的学号、课程号、学生成绩；     select DISTINCT s1.course_id,s2.course_id,s1.num,s2.num from score as s1, score as s2 where s1.num = s2.num and s1.course_id != s2.course_id;   35、查询每门课程成绩最好的前两名；       select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join     (     select         sid,         (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0,1) as first_num,         (select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 1,1) as second_num     from         score as s1     ) as T     on score.sid =T.sid     where score.num <= T.first_num and score.num >= T.second_num   36、检索至少选修两门课程的学生学号；     select student_id from score group by student_id having count(student_id) > 1   37、查询全部学生都选修的课程的课程号和课程名；     select course_id,count(1) from score group by course_id having count(1) = (select count(1) from student);   38、查询没学过“叶平”老师讲授的任一门课程的学生姓名；     select student_id,student.sname from score     left join student on score.student_id = student.sid     where score.course_id not in (         select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '张磊老师'     )     group by student_id   39、查询两门以上不及格课程的同学的学号及其平均成绩；       select student_id,count(1) from score where num < 60 group by student_id having count(1) > 2   40、检索“004”课程分数小于60，按分数降序排列的同学学号；     select student_id from score where num< 60 and course_id = 4 order by num desc;   41、删除“002”同学的“001”课程的成绩；     delete from score where course_id = 1 and student_id = 2