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数据库经典习题,

时间:2017-09-12 20:56:57      阅读:430      评论:0      收藏:1      [点我收藏+]

标签:exist   teacher   姓名   com   lap   inno   sed   tran   begin   

技术分享
/*
 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, , 1, 理解), (2, , 1, 钢蛋), (3, , 1, 张三), (4, , 1, 张一), (5, , 1, 张二), (6, , 1, 张四), (7, , 2, 铁锤), (8, , 2, 李三), (9, , 2, 李一), (10, , 2, 李二), (11, , 2, 李四), (12, , 3, 如花), (13, , 3, 刘三), (14, , 3, 刘一), (15, , 3, 刘二), (16, , 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;
mysql习题代码
技术分享
1、查询所有的课程(course)的名称以及对应的任课(teacher)老师姓名
select cname,tname from course left join teacher on course.teacher_id=teacher.tid;


2、查询学生表(student)中男女生各有多少人
select gender,count(sid) from student group by gender;


3、查询物理(course)成绩(score)等于100的学生(student)的姓名
select sname from student where sid in(select student_id from score where course_id=(select cid from course where cname=物理)and num=100);


4、查询平均成绩(score)大于八十分的同学(student)的姓名和平均成绩
select student.sname,t1.平均成绩 from student inner join
(select student_id,avg(num) 平均成绩 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(student)的学号,姓名,选课数(course),总成绩(score)
select student.sid,student.sname 姓名,选课数,总成绩 from student inner join 
(select student_id,count(course_id) 选课数,sum(num) 总成绩 from score group by student_id) as t1
on student.sid=t1.student_id;


6、 查询姓李(teacher)老师的个数
select count(tid) 李姓老师 from teacher where tname like 李%;


7、 查询没有报李平(teacher)老师课(course)的学生(student)姓名
select sname from student where sid not in
(select distinct student_id from score where course_id in(select cid from course where teacher_id=(select tid from teacher where tname=李平老师)));


8、 查询物理课程(course)比生物课程高(score)的学生(student)的学号
select t1.student_id from 
(select student_id,num from course inner join score on score.course_id=course.cid where course.cname=物理) as t1
inner join
(select student_id,num from course inner join score on score.course_id=course.cid where course.cname=生物) as t2
on t1.student_id=t2.student_id
where t1.num>t2.num;


9、 查询没有同时(score)选修物理课程(course)和体育课程的学生(student)姓名
select sname from student where sid in
(select student_id from score inner join course on course.cname in(物理,体育)and course.cid=score.course_id group by student_id having count(course_id)=1);


10、查询挂科(score)超过两门(包括两门)的学生(student)姓名和班级(course)

select t2.sname,class.caption from    #民投资,班级信息
(select sname,class_id from student inner join  #名字班级挂科超过两次
(select student_id from score where num<60 group by student_id having count(course_id)>=2) as t1  #挂科2次
on student.sid=t1.student_id) as t2
inner join class 
on class.cid = t2.class_id

11 、查询选修了所有课程的学生姓名
select sname from student inner join
(
select student_id from score group by student_id having count(course_id) = (select count(cid) from course)
) t1
on t1.student_id = student.sid
;

12、查询李平老师(teacher)教的课程的所有成绩(score)记录
 select student_id,course_id,num from score inner join (select cid from course inner join teacher on teacher.tname=李平老师 and teacher.tid=course.teacher_id) as t1
 on t1.cid=score.course_id;
 
 
#13、查询全部(student)学生都选修了的课程号和课程名(course)
select course.cid,course.cname from course inner join
(select course_id from score group by course_id having count(student_id)=(select count(sid) from student))as t1
on t1.course_id=course.cid;


#14、查询每门课程(course)被选修的次数(score)
select course.cname,选修人数 from course inner join     #课程名,人数(次数)
(select course_id,count(student_id) 选修人数 from score group by course_id) as t1   #课程ID学生总人数
on t1.course_id=course.cid;


15、查询只选修了一门课程(course)的学生(student)姓名和学号
select sid,sname from student inner join (select student_id from score group by student_id having count(course_id)=1) as t1               #一门课程    student_id只选择一门课程的学生
on t1.student_id=student.sid;


#16、查询所有学生(student)考出的成绩(score)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(score)大于85的学生(student)姓名和平均成绩
select student.sname,平均成绩 from student inner join
(select student_id,avg(num) 平均成绩 from score group by student_id having avg(num)>85) as t1 
on student.sid=t1.student_id;


18、查询生物(course)成绩(score)不及格的学生(student)姓名和对应生物分数
select student.sname,t1.num from student inner join
(select student_id,num from score where course_id=(select cid from course where cname=生物) and num<60) as t1  #s生物不及格的人的学生ID和姓名
on student.sid=t1.student_id;


19、查询在所有选修了李平(teacher)老师课程的学生(student)中,这些课程(李平老师的课程,不是所有课程)平均成绩(num)最高的学生姓名
select sname from student where sid=                   #最终筛选
(select student_id from score where course_id in     #学生的sid在李平老师的课程
(select cid 李平老师课程 from teacher inner join course on teacher.tname=李平老师 and course.teacher_id=teacher.tid)    #链接的纽带         得到李平老师课程的cid
group by student_id          #根据学生分组
order by avg(num) desc       #成绩排序
limit 1);               #去一个     最高的成绩


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
答案部分

 

数据库经典习题,

标签:exist   teacher   姓名   com   lap   inno   sed   tran   begin   

原文地址:http://www.cnblogs.com/52-qq/p/7511890.html

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