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
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;
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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
|