一、练习。导入下面sql执行语句
/*数据导入:
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;
init.sql
导入方法。路径不支持中文
#准备表、记录
mysql>create database db1;
mysql>use db1;
mysql> source /root/init.sql
生成如下表
二、练习题部分
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11、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
View Code
三、参考答案。
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。有问题
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
分析
teacher表,李平老师(tid= 2)
course表,发现李平老师教 物理, cid= 2score表,发现 cid=2的就是物理,寻找有物理分的学生idselect sname from student where sid not in(select student_id from score where course_id = 2);
8、 查询物理课程比生物课程高的学生的学号
select t1.student_id from(select student_id,num from score where course_id =(select cid from course where cname = "物理")
)ast1
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_idwhere 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
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
select student.sname,class.caption fromstudent
inner join
(select student_id from score where num < 60group by student_id
having count(course_id)>= 2)ast1
inner joinclass 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
select student.sname from student where sid in(select student_id fromscore
group by student_id
having count(course_id)=(select count(cid) fromcourse)
);
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 fromcourse
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
select * fromstudent
left join
(select student_id,group_concat(cname) fromscore
inner join
course on score.course_id=course.cid
group by student_id)ast1
on student.sid= t1.student_id;
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
select sid,sname from student where sid in(select student_id fromscore
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)
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 fromstudent
inner join
(select student_id,avg(num) as avg_num fromscore
group by student_id
having avg(num)> 85)as 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 fromscore
left join course on score.course_id=course.cid
left join student on score.student_id=student.sidwhere 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 fromcourse
inner join
teacher on course.teacher_id=teacher.tidwhere teacher.tname = '李平老师')
group by student_id
order by avg(num) desc
limit1);
20、查询每门课程成绩最好的前两名学生姓名
第一步:求出每门课程的课程course_id,与最高分数first_num
第二步: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 fromscore
inner join
(select course_id,max(num) first_num fromscore
group by course_id
)as t on score.course_id =t.course_idwhere score.num
group by course_id;
第三步: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 fromscore
group by course_id
)ast1
inner join
(select score.course_id,max(num) second_num fromscore
inner join
(select course_id,max(num) first_num fromscore
group by course_id)as t on score.course_id =t.course_idwhere score.num
group by course_id
)as t2 on t1.course_id = t2.course_id;
调整
select score.student_id,t3.course_id,t3.first_num,t3.second_num fromscore
inner join
(select t1.course_id,t1.first_num,t2.second_num from(select course_id,max(num) first_num fromscore
group by course_id
)ast1
inner join
(select score.course_id,max(num) second_num fromscore
inner join
(select course_id,max(num) first_num fromscore
group by course_id)as t on score.course_id =t.course_idwhere score.num
group by course_id
)as t2 on t1.course_id =t2.course_id
)as t3 on score.course_id =t3.course_idwhere score.num >=t3.second_num
and score.num<=t3.first_num
order by course_id;
检验
21、查询不同课程但成绩相同的学号,课程号,成绩