mysql 练习题
导出现有数据库数据：
- C:\Users\Administrator>mysqldump -u root db1>D:\agon\db1.sql -p #结构+数据
- mysqldump -u用户名 -d 数据库名称 >导出文件路径 -p #结构
导入现有数据库：
- mysqldump -u root 数据库名称 < 文件路径 -p #箭头是重点
/*
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 ('', '三年二班'), ('', '三年三班'), ('', '一年二班'), ('', '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 ('', '生物', ''), ('', '物理', ''), ('', '体育', ''), ('', '美术', '');
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 ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', ''), ('', '', '', '');
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 ('', '男', '', '理解'), ('', '女', '', '钢蛋'), ('', '男', '', '张三'), ('', '男', '', '张一'), ('', '女', '', '张二'), ('', '男', '', '张四'), ('', '女', '', '铁锤'), ('', '男', '', '李三'), ('', '男', '', '李一'), ('', '女', '', '李二'), ('', '男', '', '李四'), ('', '女', '', '如花'), ('', '男', '', '刘三'), ('', '男', '', '刘一'), ('', '女', '', '刘二'), ('', '男', '', '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 ('', '张磊老师'), ('', '李平老师'), ('', '刘海燕老师'), ('', '朱云海老师'), ('', '李杰老师');
COMMIT; SET FOREIGN_KEY_CHECKS = 1;
表结构和数据
C:\Users\Administrator>mysqldump -u root db1>D:\agon\db1.sql -p 转储命令
select * from course;
SELECT teacher_id,count(cname) from course group by teacher_id;
select * from course left join teacher on course.teacher_id=teacher.tid;
SELECT * from student left join class on student.class_id=class.cid;
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：获取所有生物课程的人(学号,成绩)-临时表
获取所有生物课程的人(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
然后在进行筛选
select A.student_id from
(select student_id,num from score LEFT JOIN course on score.course_id=course.cid WHERE cname='生物') as A
left join
(select student_id,num 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；
select B.student_id,student.sname,b.avv from
(SELECT student_id,avg(num) as avv from score GROUP BY student_id having avg(num)>60) as B
left join student on B.student_id=student.sid; 4、查询所有同学的学号、姓名、选课数、总成绩；student.sid,count(student.sid)
select student_id,count(student_id),sum(num),student.sname from score
left join student on score.student_id=student.sid
group by student_id; 5、查询姓“李”的老师的个数；
select count(tname) from teacher where tname like'李%'; 6、查询没学过“叶平”老师课的同学的学号、姓名；
获取李平老师课程ID
select cid from course left join teacher on course.teacher_id=teacher.tid where tname='李平老师';
查询所有学过老师课的学生的id,
select student_id from score where course_id in (2,4) group by student_id;
select * from tb12 where id in (select id from tb11) 这个是格式 select student.sid,student.sname from student where sid not in
(select student_id from
(select student_id from score where course_id in (select cid from course left join teacher on course.teacher_id=teacher.tid where tname='李平老师') group by student_id)as B
);
7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名；
找到所有选了001课程的同学的id和所有选了002课程的同学的id 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); 不加having也行 8、查询学过“叶平”老师所教的所有课的同学的学号、姓名；
先找李平老师教过的课的id
select cid from course LEFT JOIN teacher on course.teacher_id=teacher.tid where tname like '李平%';
通过cid找到对应的sid
select student_id,sname from student INNER JOIN
(SELECT student_id from score where course_id in (select cid from course LEFT JOIN teacher on course.teacher_id=teacher.tid where tname like '李平%'))as B
on B.student_id=student.sid GROUP BY student_id HAVING COUNT(student_id)>1; 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；
先找到002课程的所有成绩和student_id在找到001的num和student_id
select student_id,course_id,num from score where course_id=2;
select student_id,course_id,num from score where course_id=1;
这里有个坑，我操操操，第57行如果有A.num和B.num的话会报重复列的错，
所以够用就行，不要多谢闲的蛋疼
SELECT student.sid,student.sname from
(
select * from
(select student_id,course_id,num from score where course_id=2)as A
INNER JOIN
(select student_id,course_id,num from score where course_id=1)as B
on A.student_id=B.student_id where A.num<B.num
)as C
INNER JOIN student
on C.student_id=student.sid; 10、查询有课程成绩小于60分的同学的学号、姓名； SELECT student.sid,sname,num FROM score
INNER JOIN student on score.student_id=student.sid
where num<60; 11、查询没有学全所有课的同学的学号、姓名；
SELECT A.student_id,count(A.student_id),sname from
(SELECT student_id,course_id from score)as A
LEFT JOIN student on A.student_id=student.sid
group by A.student_id HAVING count(A.student_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；
先找到学好为001的学生的课程ID 124
限制死了，就算你有三门也是124里面的三门，注意一下
select sid,sname from
(SELECT student_id,course_id from score where student_id!=1 and course_id in (1,2,4))as A
LEFT JOIN student on A.student_id=student.sid GROUP BY sid; 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；
也是限制死了，从select * from score WHERE course_id in (1,2,4) 分组后计数最多为3
但是可能人家是选了四门课，这是注意点
SELECT student.sid,student.sname from student
INNER JOIN
(
select student_id,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(1)=(SELECT count(course_id) from score WHERE student_id=1)
)as A
on student.sid=A.student_id; 14、查询和“001”号的同学学习的课程完全相同的其他同学学号和姓名；
先找到和001一样都学过3门的学生id
在通过course_id将表约束为只有1,2,4三门的学生成绩表，通过这个表分组筛选出个数为3的student_id明显没有符合条件的
select student_id from score where student_id in
(select student_id from score where student_id!=1 GROUP BY student_id having count(course_id)=3)
and course_id in(1,2,4) GROUP BY student_id having count(1)=3; select student_id from score where student_id in
(
select student_id from score where student_id !=1 GROUP BY student_id HAVING count(1) = (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(1) = (select count(1) from score where student_id = 1) 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 LEFT JOIN teacher on course.teacher_id=teacher.tid WHERE 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；
用到insert into 表名(a,b,c) select a,b,c from 表名 // 如果bc不是表的列明则bc必须是单值
先找到选了002课程的所有同学的id
主要是讲新的插入姿势
insert into score(student_id,course_id,num)
select student_id,2,(select avg(num) from score where course_id=2) 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；
知识点把列的内容横着放，涉及到子查询，每个学生ID对应三门成绩，还需把所有的null设置为0
select
student_id,
(select num from score as s2 where s2.student_id=s1.student_id and course_id=1)as 语文,
(select num from score as s2 where s2.student_id=s1.student_id and course_id=3)as 数学,
(select num from score as s2 where s2.student_id=s1.student_id and course_id=4)as 英语,
avg(if(isnull(s1.num),0,s1.num)),
avg(num)
from score as s1 GROUP BY student_id ORDER BY avg(if(isnull(s1.num),0,s1.num)) 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),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；
知识点：select case when num<60 then 0 else num end from score; case when..THEN..ELSE..END 可直接用于查
大于60分的=1，小于60分的=0，二者求和后就是及格人数，用及格人数/选该科目的总人数 就是及格率 select
course_id,avg(num),
sum(case when score.num>60 then 1 else 0 end)/count(1)*100 as percent
from score GROUP BY course_id ORDER BY avg(num) asc,percent desc; 20、课程平均分从高到低显示（现实任课老师）；
知识点：avg(if(isnull(score.num),0,score.num)) 如果为null则为0否则为null.num select course_id,teacher.tname,avg(if(isnull(score.num),0,score.num)) from score
LEFT JOIN course on score.course_id=course.cid
LEFT JOIN teacher on course.teacher_id=teacher.tid
GROUP BY course_id ORDER BY avg(if(isnull(score.num),0,score.num)) desc; 21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
考虑成绩并列的话，应该先按成绩分组，这样如果有3个100分则他们都是第一名
select
course_id,
(select num from score as s1 WHERE s1.course_id=s2.course_id GROUP BY num ORDER BY num desc LIMIT 0,1)as A,
(select num from score as s1 WHERE s1.course_id=s2.course_id GROUP BY num ORDER BY num desc LIMIT 1,1)as B,
(select num from score as s1 WHERE s1.course_id=s2.course_id GROUP BY num ORDER BY num desc LIMIT 2,1)as C
from score as s2 GROUP BY course_id; 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_id,student.sname from score
LEFT JOIN student on score.student_id=student.sid
GROUP BY student_id HAVING COUNT(course_id)=1 24、查询男生、女生的人数；
select gender,COUNT(1) from student GROUP BY gender; 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) from student GROUP BY sname HAVING count(1)>1; 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)) from score
GROUP BY course_id ORDER BY avg(if(isnull(num),0,num)) 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 HAVING avg(if(isnull(num),0,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.sid,sname from score
LEFT JOIN course on score.course_id=course.cid
LEFT JOIN student on score.student_id=student.sid
where cname='生物' and num<60 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 score.student_id=student.sid
where course_id=3 and num>80 31、求选了课程的学生人数
select count(distinct student_id) from score; select count(1) from
(select student_id 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 course on score.course_id=course.cid
LEFT JOIN student on score.student_id=student.sid
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 course.cid=score.course_id
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 * from teacher as s1,teacher as s2; 5*5=25 笛卡尔机原理
select s1.student_id,s1.course_id,s1.num from score as s1,score as s2
where s1.sid!=s2.sid and s1.course_id!=s2.course_id and s1.num=s2.num; 35、查询每门课程成绩最好的前两名；
SELECT
course_id,
(select num from score as s1 where s1.course_id=s2.course_id GROUP BY num ORDER BY num desc LIMIT 0,1)as A,
(select num from score as s1 where s1.course_id=s2.course_id GROUP BY num ORDER BY num desc LIMIT 1,1)as B
from score as s2 GROUP BY course_id; 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(course_id)>1; 37、查询全部学生都选修的课程的课程号和课程名；
把score以课程号分组加个条件：count(1)=student表中的行数===》这样拿到课程id，连表course拿到课程名
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；
先找到李平老师教过的所有课的学生的id SELECT sid,sname from student where sid not in
(
select student_id from score WHERE course_id 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),avg(if(ISNULL(num),0,num)) 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 student_id=2 and course_id=1;
mysql总结
# SQL 语句的格式
SELECT DISTINCT <select_list>
FROM <left_table>
<join_type> JOIN <right_table>
ON <join_condition>
WHERE <where_condition>
GROUP BY <group_by_list>
HAVING <having_condition>
ORDER BY <order_by_condition>
LIMIT <limit_number> # SQL 语句的执行顺序
(1) FROM <left_table>
(2) ON <join_condition>
(3) <join_type> JOIN <right_table>
(4) WHERE <where_condition>
(5) GROUP BY <group_by_list>
(6) HAVING <having_condition>
(7) SELECT
(8) DISTINCT <select_list>
(9) ORDER BY <order_by_condition>
(10) LIMIT <limit_number>
mysql 练习题的更多相关文章
- MySQL练习题
MySQL练习题 一.表关系 请创建如下表,并创建相关约束 二.操作表 1.自行创建测试数据 2.查询“生物”课程比“物理”课程成绩高的所有学生的学号: 3.查询平均成绩大于60分的同学的学号和平均成 ...
- MySQL练习题参考答案
MySQL练习题参考答案 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: 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[ ...
- s15day12作业:MySQL练习题参考答案
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 ...
- Python/ MySQL练习题（一）
Python/ MySQL练习题(一) 查询“生物”课程比“物理”课程成绩高的所有学生的学号 SELECT * FROM ( SELECT * FROM course LEFT JOIN score ...
- python/MySQL练习题(二)
python/MySQL练习题(二) 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 ...
- python 全栈开发，Day65(MySQL练习题,参考答案)
一.MySQL练习题 一.表关系 请创建如下表,并创建相关约束 二.操作表 1.自行创建测试数据 2.查询“生物”课程比“物理”课程成绩高的所有学生的学号.ps:针对的是自己的生物成绩比物理成绩高,再 ...
- mysql 练习题答案
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 ...
- mysql练习题练习
1.数据库是按照原文制作的,表格结构一样具体存储的数据有些差异 原文地址:MySQL练习题 原答案地址:MySQL练习题参考答案 2.查询“生物”课程比“物理”课程成绩高的所有学生的学号: selec ...
- MySQL练习题及答案（复习）
新建一个叫做 review 的数据库,将测试数据脚本导进去.(可以使用Navicat查询功能) /* Navicat MySQL Data Transfer Source Server : DB So ...
- mysql练习题-查询同时参加计算机和英语考试的学生的信息-遁地龙卷风
(-1)写在前面 文章参考http://blog.sina.com.cn/willcaty. 针对其中的一道练习题想出两种其他的答案,希望网友给出更多回答. (0) 基础数据 student表 +-- ...
随机推荐
- cesium Animation显示系统时间
var d = new Date(); var hour = 0 - d.getTimezoneOffset(); viewer.animation.viewModel.timeFormatter = ...
- java安全框架shiro(一)
第一个简单的案例 ,通过读取.ini文件的方式模拟登陆, 1.通过Factory工厂的getInstance()方法来获取SecurityManager的实例,实例化Factory需要一个ini文件的 ...
- MySQL数据库学习三 数据库对象和基本操作
3.1 数据库和数据对象 1. 系统数据库 information_schema:主要存储系统中的一些数据库对象信息,如用户列表信息.列信息.权限信息.字符集信息和分区信息等. performance ...
- 网络通信 --> 消息队列
消息队列 消息队列提供了一种从一个进程向另一个进程发送一个数据块的方法.可以通过发送消息来避免命名管道的同步和阻塞问题.但是消息队列与命名管道一样,每个数据块都有一个最大长度的限制. Linux用宏M ...
- MyBatis-plus 代码生成器
1.添加pom文件依赖 <!-- Mybatis-Plus 自动生成实体类--> <dependency> <groupId>com.baomidou</gr ...
- Alpha第一天
Alpha第一天 听说 031502543 周龙荣(队长) 031502615 李家鹏 031502632 伍晨薇 031502637 张柽 031502639 郑秦 1.前言 任务分配是VV.ZQ. ...
- Beta阶段报告
Beta版测试报告 1. 在测试过程中总共发现了多少Bug?每个类别的Bug分别为多少个? BUG名 修复的BUG 不能重现的BUG 非BUG 没能力修复的BUG 下个版本修复 url乱码 √ 手机端 ...
- Beta冲刺NO.1
Beta冲刺 第一天 1. 昨天的困难 由于今天还是第一天,所以暂时没有昨天的困难. 2. 今天解决的进度 潘伟靖: 对代码进行了review 1.将某些硬编码改为软编码 2.合并了一些方法,简化代码 ...
- 2018c语言第1次作业
6-1 计算两数的和与差 1.设计思路 (1)主要描述题目算法 第一步:把两个数的加减法分别赋给psum和pdiff. 第二步:通过psum和pdiff的地址把值传回主函数. (2)流程图.(无) 2 ...
- python functools.lru_cache做备忘功能
import time import functools def clock(func): @functools.wraps(func)#还原被装饰函数的__name__和__doc__属性 def ...