Mysql练习
表关系
请创建如下表，并创建相关约束
答案如下：
create table class(
cid bigint not null auto_increment primary key,
caption char(20)
)ENGINE=INNODB DEFAULT CHARSET=utf8;
insert into class(caption) values("三年二班"),("一年三班"),("三年一班");
create table teacher(
tid int not null auto_increment primary key,
tname char(20)
)engine = innodb default charset=utf8;
insert into teacher(tname) values ("波多"),("苍空"),("饭岛");
create table student(
sid bigint not null auto_increment primary key,
sname char(20),
gender char(5),
class_id int,
constraint fk_student_class_id_class_cid foreign key (class_id) references class(cid)
)engine=innodb default charset=utf8;
insert into student(sname,gender,class_id) values("钢蛋","女","1"),("铁锤","女","1"),("山炮","男","2");
create table course(
cid bigint not null auto_increment primary key,
cname char(20),
teach_id int,
constraint fk_course_teach_id_teacher_tid foreign key (teach_id) references teacher(tid)
)engine=innodb default charset=utf8;
insert into course(cname,teach_id) values("生物",1),("体育",1),("物理",2);
create table score(
sid bigint not null auto_increment primary key,
student_id int,
course_id int,
number int,
constraint fk_score_student_id_student_sid foreign key (student_id) references student(sid),
constraint fk_score_corse_id_course_sid foreign key (student_id) references course(cid)
)engine = innodb default charset=utf8;
insert into score(student_id,course_id,number) values (1,1,60),(1,2,59),(2,2,100);
操作表
1、自行创建测试数据
2、查询“生物”课程比“物理”课程成绩高的所有学生的学号；
3、查询平均成绩大于60分的同学的学号和平均成绩；
4、查询所有同学的学号、姓名、选课数、总成绩；
5、查询姓“李”的老师的个数；
6、查询没学过“李平”老师课的同学的学号、姓名； --------------先得知道李平老师教哪些课
7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；
8、查询学过“李平”老师所教的所有课的同学的学号、姓名；
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；
10、查询有课程成绩小于60分的同学的学号、姓名；
11、查询没有学全所有课的同学的学号、姓名；
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；
13、查询至少学过学号为“001”同学所选课程中任意一门课的其他同学学号和姓名；
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；
15、删除学习“叶平”老师课的SC表记录；
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；
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；
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；
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；
20、课程平均分从高到低显示（现实任课老师）；
21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
22、查询每门课程被选修的学生数；
23、查询出只选修了一门课程的全部学生的学号和姓名；
24、查询男生、女生的人数；
25、查询姓“张”的学生名单；
26、查询同名同姓学生名单，并统计同名人数；
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；
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；
29、查询课程名称为“数学”，且分数低于60的学生姓名和分数；
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；
31、求选了课程的学生人数
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；
33、查询各个课程及相应的选修人数；
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；
35、查询每门课程成绩最好的前两名；
36、检索至少选修两门课程的学生学号；
37、查询全部学生都选修的课程的课程号和课程名；
38、查询没学过“叶平”老师讲授的任一门课程的学生姓名；
39、查询两门以上不及格课程的同学的学号及其平均成绩；
40、检索“004”课程分数小于60，按分数降序排列的同学学号；
41、删除“002”同学的“001”课程的成绩；
数据的导入导出
导出现有数据库数据：
- mysqldump -u用户名 -p密码 数据库名称 >导出文件路径 # 结构+数据
- mysqldump -u用户名 -p密码 -d 数据库名称 >导出文件路径 # 结构
导入现有数据库数据：
- mysqldump -uroot -p密码 数据库名称 < 文件路径
/*
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;
表结构和数据
相关
distinct 去重
select distinct student_id from score where num<60;
等于
select student_id from score where num<60 group by student_id;
case ... when ... end
case when score.num > 60 then 1 else 0 END；--------如果score.num大于60，则score这列等于1，否则等于0.
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;
# 多重判断
select case sname
when '张一' then 'good'
when '张二' then 'top'
end
as AU
from student;
# 多重判断另一种写法
select
case when author='Felix' then 'good'
when author='Tom' then 'top'
when author='Bob' then 'down'
else 'do not know'
end
as AU
from felix_test;
# if
if(score.num>60, 1, 0)
select course_id, avg(num) as avgnum,sum(if(score.num>60, 1, 0))/count(1)*100 as percent from score group by course_id order by avgnum asc,percent desc;
# ifnull，注意参数顺序。为null取第二个参数
select course_id,avg(if(isnull(num), 0 ,num)) as avg from score group by course_id order by avg asc,course_id desc;
select course_id,avg(ifnull(num,0)) as avg from score group by course_id order by avg asc,course_id desc;
第二种写法
select * from score as a, score as b where a.student_id = b.student_id and a.course_id="1" and b.course_id=2 and a.num < b.num
解答
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);
where sw > ty 和 where A.num > B.num 相等
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
3.1 额外加一个任务，还需要拿到学生的姓名。（此时先需要连表操作）
select student_id,avg(num),sname from score
left join student on score.student_id = student.sid
group by student_id having avg(num) > 60
这里的操作是先连表，再组合取数据。
select * from (select student_id,avg(num) from score group by student_id HAVING avg(num)>60) as T
left join student on T.student_id = student.sid;
# 这是第二种操作，但是更复杂。
select T.student_id,T.a,student.sname from (select student_id,avg(num) as a from score group by student_id HAVING avg(num)>60) as T
left join student on T.student_id = student.sid;
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 student.sid,student.sname FROM student WHERE sid NOT IN (
SELECT student_id FROM score WHERE course_id IN (
SELECT course.cid FROM course LEFT JOIN teacher ON course.teacher_id=teacher.tid WHERE teacher.tname="李平老师") GROUP BY student_id)
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 score.student_id,student.sname from score
left join student on score.student_id = student.sid
where course_id=1 or course_id=2 group by student_id HAVING count(course_id)>1;
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；
SELECT B.student_id,student.sname from (
SELECT student_id FROM score WHERE course_id IN (
SELECT cid FROM course LEFT JOIN teacher ON course.teacher_id=teacher.tid WHERE teacher.tname="李平老师") GROUP BY student_id HAVING count(course_id)=(
SELECT count(cid) FROM course LEFT JOIN teacher ON course.teacher_id=teacher.tid WHERE teacher.tname="李平老师")) AS B LEFT JOIN student ON B.student_id=student.sid;
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2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；
方式一：条件判断 case when .. then ，when .. then,end
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;
case when score.num > 60 then 1 else 0 END；--------如果score.num大于60，则score这列等于1，否则等于0.
方式二：条件判断，if
select course_id, avg(num) as avgnum,sum(if(score.num>60, 1, 0))/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
if(isnull(score.num),0,score.num);-----------三元运算，如果等于null则改成0，不是null则是自己。防止统计sum什么的时候由于null报错；
21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
SELECT
T1.*
FROM
score T1
WHERE
( SELECT COUNT( 1 ) FROM score WHERE T1.course_id = course_id AND T1.num < num ) < 3
ORDER BY
course_id,
num DESC;
我们要找出集合中最大的三个数，就要拿 T1 的每个元素和 T2 比较， 如果 T1 的某个元素在 T2 的集合中只找到三个以下比它大的元素， 说明这个元素是前三名之一
这写法，会把成绩相同的所有人视为1个人
SELECT
T1.*
FROM
score T1
LEFT JOIN ( SELECT DISTINCT course_id, num FROM score ) T2 ON T1.course_id = T2.course_id
AND T1.num < T2.num
GROUP BY
student_id,
course_id,
num
HAVING
COUNT( 1 ) < 3
ORDER BY
course_id,
num DESC
除了DISTINCT ，其他跟方式一一样。方式一加DISTINCT也可以
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；
方式一
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
方式二
select (select count(1) from student where gender='男') as man,(select count(1) from student where gender='女') as woman;
方式三
select (select count(1) from student where gender='男') as man,(select count(1) from student where gender='女') as woman from DUAL;
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 DISTINCT score.student_id,student.sname from score left join student on score.student_id=student.sid 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;
# mark：比较有意思的写法
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
