mysql练习题
/* 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
练习题与参考答案:
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:1、查询生物的所有学生的成绩学号
2、查询物理的所有学生的成绩学号
3、将两个结果聚合到一起，在做一次判断
select A.student_id from
(select score.student_id,score.num,course.cname from score LEFT JOIN course on score.course_id = course.cid where cname = '生物') as A
LEFT JOIN
(select score.student_id,score.num,course.cname from score LEFT JOIN course on score.course_id = course.cid where cname = '物理') as B
on
A.student_id = B.student_id where A.num > B.num
3、查询平均成绩大于60分的同学的学号和平均成绩；
1、根据学生id进行分组,使用avg获取平均值,通过having进行平均值判断
select student.sname,A.num from (select student_id,avg(num) as num from score group by student_id having avg(num) > 60 ) as A
LEFT JOIN
student on A.student_id = student.sid
4、查询所有同学的学号、姓名、选课数、总成绩；
1、通过score表查看所有人的选课数和总成绩
2、在和student表做一次join聚合
select student.sid,student.sname,A.times,A.num
from
(select student_id,count(student_id) as times,sum(num)as num from score group by student_id ) as A
LEFT JOIN
student on A.student_id = student.sid
1、先聚合在查询
另外一种方法
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；
1、先查老师的id 在查非李平老师id的学生,在做一次去重
select sid,sname from student where sid not in (
(select distinct student_id from score LEFT JOIN course on score.course_id = course.cid
WHERE teacher_id in (select tid from teacher where tname = '李平老师')))
另外一种思路：
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；
1、先查到学生学001又学002课程的人,根据学生分组课程数量等于2
2、在和学生表结合取出学生id和学生名字
select student_id,student.sname from
(select student_id,count(student_id) as total from score
WHERE course_id in (1,2) GROUP BY student_id HAVING total = 2) as A
LEFT JOIN student on student.sid = A.student_id
另外一种写法:
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；
select sid,sname from student where sid in (
(select distinct student_id from score LEFT JOIN course on score.course_id = course.cid
WHERE teacher_id in (select tid from teacher where tname = '李平老师')))
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；
select A.student_id from
(select score.student_id,score.num,student.sname from score LEFT JOIN student on score.student_id = student.sid where course_id = 1) as A
LEFT JOIN
(select score.student_id,score.num,student.sname from score LEFT JOIN student on score.student_id = student.sid where course_id = 2) as B
on
A.student_id = B.student_id where A.num > B.num
10、查询有课程成绩小于60分的同学的学号、姓名；
1、先查询学生小于60分的人员 在和学生表进行关联
select student.sid,student.sname from (select student_id from score where num < 60 group by student_id)
as A left join student on A.student_id = student.sid
2、第二种方法
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；
1、先在score里查询没有学全的同学，在student表里进行一次查询即可
select sid,sname from student where sid in (select student_id from score GROUP BY student_id having count(student_id) < (select count(1) from course) )
另外一种方法
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；
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 sid,sname from student where sid in
(select student_id from score where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id)
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”同学所选课程中任意一门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 sid,sname from student where sid in
(select student_id 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(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 != 3 group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 3)
) and course_id in (select course_id from score where student_id = 3) group by student_id HAVING count(course_id) = (select count(1) from score where student_id = 3)
15、删除学习“叶平”老师课的SC表记录；
delete from score where course_id in (select cid from course where teacher_id in (select tid from teacher where teacher.tname = '李平老师') )
另外一种方法
delete 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'
)
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
)
53 2 2 65
54 13 2 65
55 14 2 65
56 15 2 65
57 16 2 65
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 asc
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.cname,A.max_num,A.min_num from (select course_id, max(num) as max_num, min(num) as min_num from score group by course_id) as A left join course on A.course_id = course.cid
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（显示任课老师）
1、先查询平均分,在进行连表,一次排序
select teacher.tname,A.avgnum from (select course_id,avg(num) as avgnum from score group by course_id) as A LEFT JOIN course on A.course_id= course.cid
LEFT JOIN teacher on course.teacher_id = teacher.tid ORDER BY A.avgnum desc
另外一种方法
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) --不清楚怎么实现的
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(student_id) from score GROUP BY course_id
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.sname,student.sid from (select student_id from score group by student_id HAVING count(course_id) = 1 ) as A
LEFT JOIN student on A.student_id = student.sid
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 * 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 * from (select course_id,avg(num) as num from score group by course_id ) as A ORDER BY A.num asc
另外一种写法
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(num) from score left JOIN student on score.student_id = student.sid group by student_id HAVING avg(num) > 85
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_id,sname,num from score LEFT JOIN student on score.student_id = student.sid
where course_id in (select cid from course where cname = '生物') and num < 60
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 student_id,sname from score LEFT JOIN student on student.sid = score.student_id where course_id = 3 and num > 80
31、求选了课程的学生人数
select count(student_id) from (select student_id from score where course_id in (1,2,3,4) 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,sname from score left join student on student.sid = score.student_id 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 = '李平老师'
)
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 * from score where course_id = 2 and num < 60 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 = 001
http://www.cnblogs.com/wupeiqi/articles/5729934.html
http://www.cnblogs.com/wupeiqi/articles/5748496.html
