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SQL练习
阅读量:4633 次
发布时间:2019-06-09

本文共 10413 字,大约阅读时间需要 34 分钟。

 

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

准备数据:init.sql文件内容:

1 /*  2      数据导入:  3      Navicat Premium Data Transfer  4   5      Source Server         : localhost  6      Source Server Type    : MySQL  7      Source Server Version : 50624  8      Source Host           : localhost  9      Source Database       : sqlexam 10  11      Target Server Type    : MySQL 12      Target Server Version : 50624 13      File Encoding         : utf-8 14  15      Date: 10/21/2016 06:46:46 AM 16     */ 17  18     SET NAMES utf8; 19     SET FOREIGN_KEY_CHECKS = 0; 20  21     -- ---------------------------- 22     --  Table structure for `class` 23     -- ---------------------------- 24     DROP TABLE IF EXISTS `class`; 25     CREATE TABLE `class` ( 26       `cid` int(11) NOT NULL AUTO_INCREMENT, 27       `caption` varchar(32) NOT NULL, 28       PRIMARY KEY (`cid`) 29     ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8; 30  31     -- ---------------------------- 32     --  Records of `class` 33     -- ---------------------------- 34     BEGIN; 35     INSERT INTO `class` VALUES ('1', '三年二班'), ('2', '三年三班'), ('3', '一年二班'), ('4', '二年九班'); 36     COMMIT; 37  38     -- ---------------------------- 39     --  Table structure for `course` 40     -- ---------------------------- 41     DROP TABLE IF EXISTS `course`; 42     CREATE TABLE `course` ( 43       `cid` int(11) NOT NULL AUTO_INCREMENT, 44       `cname` varchar(32) NOT NULL, 45       `teacher_id` int(11) NOT NULL, 46       PRIMARY KEY (`cid`), 47       KEY `fk_course_teacher` (`teacher_id`), 48       CONSTRAINT `fk_course_teacher` FOREIGN KEY (`teacher_id`) REFERENCES `teacher` (`tid`) 49     ) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8; 50  51     -- ---------------------------- 52     --  Records of `course` 53     -- ---------------------------- 54     BEGIN; 55     INSERT INTO `course` VALUES ('1', '生物', '1'), ('2', '物理', '2'), ('3', '体育', '3'), ('4', '美术', '2'); 56     COMMIT; 57  58     -- ---------------------------- 59     --  Table structure for `score` 60     -- ---------------------------- 61     DROP TABLE IF EXISTS `score`; 62     CREATE TABLE `score` ( 63       `sid` int(11) NOT NULL AUTO_INCREMENT, 64       `student_id` int(11) NOT NULL, 65       `course_id` int(11) NOT NULL, 66       `num` int(11) NOT NULL, 67       PRIMARY KEY (`sid`), 68       KEY `fk_score_student` (`student_id`), 69       KEY `fk_score_course` (`course_id`), 70       CONSTRAINT `fk_score_course` FOREIGN KEY (`course_id`) REFERENCES `course` (`cid`), 71       CONSTRAINT `fk_score_student` FOREIGN KEY (`student_id`) REFERENCES `student` (`sid`) 72     ) ENGINE=InnoDB AUTO_INCREMENT=53 DEFAULT CHARSET=utf8; 73  74     -- ---------------------------- 75     --  Records of `score` 76     -- ---------------------------- 77     BEGIN; 78     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'); 79     COMMIT; 80  81     -- ---------------------------- 82     --  Table structure for `student` 83     -- ---------------------------- 84     DROP TABLE IF EXISTS `student`; 85     CREATE TABLE `student` ( 86       `sid` int(11) NOT NULL AUTO_INCREMENT, 87       `gender` char(1) NOT NULL, 88       `class_id` int(11) NOT NULL, 89       `sname` varchar(32) NOT NULL, 90       PRIMARY KEY (`sid`), 91       KEY `fk_class` (`class_id`), 92       CONSTRAINT `fk_class` FOREIGN KEY (`class_id`) REFERENCES `class` (`cid`) 93     ) ENGINE=InnoDB AUTO_INCREMENT=17 DEFAULT CHARSET=utf8; 94  95     -- ---------------------------- 96     --  Records of `student` 97     -- ---------------------------- 98     BEGIN; 99     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');100     COMMIT;101 102     -- ----------------------------103     --  Table structure for `teacher`104     -- ----------------------------105     DROP TABLE IF EXISTS `teacher`;106     CREATE TABLE `teacher` (107       `tid` int(11) NOT NULL AUTO_INCREMENT,108       `tname` varchar(32) NOT NULL,109       PRIMARY KEY (`tid`)110     ) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;111 112     -- ----------------------------113     --  Records of `teacher`114     -- ----------------------------115     BEGIN;116     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');117     COMMIT;118 119     SET FOREIGN_KEY_CHECKS = 1;120 121 准备数据122 123 准备数据
View Code

 

从init.sql文件中导入数据

#准备表、记录mysql> create database db8;mysql> use db8;mysql> source D:\MySQL\mysql\data\db8\init.sql

   

      

 

1、查询所有的课程的名称以及对应的任课老师姓名

思路:这道题牵扯到两张表course,teacher;需要用到内连接select course.cname,teacher.tnamefrom 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

思路:题目中有'每','各'都应该想到分组查询group byselect gender,count(sid) from student group by gender;

  

3、查询物理成绩等于100的学生的姓名

思路:这道题需要用到三张表score,student,,course先查询课程号为物理的课程,然后再在成绩表中去查找成绩为100的学生sid,最后再用找到sid的去找学生姓名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、查询平均成绩大于八十分的同学的姓名和平均成绩

思路:用到表score,student先求平均成绩,然后再在学生表里查找学生姓名,最后用内连接select student_id,avg(num) as avg_num  from score group by student_id having avg(num)>80

  

select student.sname,t1.avg_num from student inner join(select student_id,avg(num) as avg_numfrom score group by student_id having avg(num)>80) as t1on student.sid=t1.student_id;

  

5、查询所有学生的学号,姓名,选课数,总成绩

思路:先查询score中的信息 然后用左链接到学生表select student_id,count(course_id) as course_num,sum(num) as total_num from score group by student_id

  

select student.sid,student.sname,t1.course_num,t1.total_numfrom  student left join  (select student_id,count(course_id) as course_num,sum(num) as total_num   from score group by student_id)as t1on student.sid=t1.student_id;

  

6、 查询姓李老师的个数

思路:只用教师表 模糊查询likeselect count(tid) from teacher where tname like '李%';

  

7、 查询没有报李平老师课的学生姓名

思路:先是course表与teacher表对应,找到报了李平老师教的课程的学生,取反(not in)select course.cid from course inner join teacher on course.teacher_id=teacher.tid where teacher.tname='李平老师';然后再在score表中找到对应的student_id,最后在去student表select distinct student_id from score where course_id in(2);

  

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

思路:先分别查询物理、生物课程的学号、成绩,然后再用内连接这两张表select student_id,num from score where course_id = (select cid from course where cname = '物理')select student_id,num from score where course_id = (select cid from course where cname = '生物')

 

select t1.student_id,t1.num,t2.numfrom (  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

思路:有三张表 course表,学生student表,成绩score表先到课程里面找到课程号,然后根据课程号去成绩表中找学号,再根据学号去学生表找学生

  

select student.sname from studentwhere 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、查询选修了所有课程的学生姓名

思路: 这道题考查3张表,学生表,课程表,成绩表先去成绩表中查找超过两门不及格的学生的学号,然后在关联到学生表,可以知道学生的姓名、学号,最后关联到班级表

  

select t1.caption,t2.sname from class as t1  inner join (    select sname, class_id from student      where sid in (select student_id from score where num < 60 group by student_id having count(sid) >= 2)      ) as t2         on t1.cid = t2.class_id;

  

11、查询李平老师教的课程的所有成绩记录

思路:三张表成绩表,课程表,教师表这道题分两步:查询所有成绩、李平老师所教的课程select * from score where course_id in (2,4);select cid from course inner join teacher on course.teacher_id=teacher.tid where teacher.tname='李平老师';

  

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'  );

  

12、查询全部学生都选修了的课程号和课程名

思路: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

 

select cid,cname from course   where cid in (    select course_id from score group by course_id having count(sid)=(select count(sid) from student)   );

  

13、查询每门课程被选修的次数

思路:通过成绩表的course_id,student_id即可查出select course_id, count(student_id) from score group by course_id;

  

14、查询之选修了一门课程的学生姓名和学号

思路:先找出只选修了一门课的student_id,然后再在学生表中找到学号,姓名select sid,sname from student  where sid in (    select student_id from score group by student_id having count(sid) = 1  );

  

15、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;

  

16、查询平均成绩大于85的学生姓名和平均成绩

思路:先去成绩表中平均成绩大于85的student_id,然后匹配到学生表即可得出答案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) > 85  ) as t1     on student.sid = t1.student_id;

  

 

转载于:https://www.cnblogs.com/xfxing/p/9333324.html

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