USE school;
-- 班级表
CREATE TABLE class(
cid TINYINT PRIMARY KEY AUTO_INCREMENT,
caption VARCHAR(20)
); INSERT INTO class(caption) VALUES("三年二班"),("一年三班"),("三年一班"); SELECT * FROM class; -- 老师表
CREATE TABLE teacher(
tid TINYINT PRIMARY KEY AUTO_INCREMENT,
tname VARCHAR(20)
); INSERT INTO teacher(tname) VALUES("波多"),("苍空"),("饭岛"); SELECT * FROM teacher; -- 学生表
CREATE TABLE student(
sid TINYINT PRIMARY KEY AUTO_INCREMENT,
sname VARCHAR(20),
gender VARCHAR(10),
class_id TINYINT,
FOREIGN KEY (class_id) REFERENCES class(cid)
); INSERT INTO student(sname,gender,class_id) VALUES
("钢蛋","女",1),
("铁锤","女",1),
("山炮","男",2); SELECT * FROM student; -- 课程表
CREATE TABLE course(
cid TINYINT PRIMARY KEY AUTO_INCREMENT,
cname VARCHAR(20),
teacher_id TINYINT,
FOREIGN KEY (teacher_id) REFERENCES teacher(tid)
);
ALTER TABLE course MODIFY cid TINYINT, DROP PRIMARY KEY;
ALTER TABLE course ADD CONSTRAINT xx FOREIGN KEY (cid) REFERENCES class(cid);
DESC course;
SHOW CREATE TABLE course;
INSERT INTO course(cname,teacher_id) VALUES
("生物",1),
("体育",1),
("物理",2); SELECT * FROM course;
-- 成绩表
CREATE TABLE score(
sid TINYINT PRIMARY KEY AUTO_INCREMENT,
student_id TINYINT,
course_id TINYINT,
number INT,
FOREIGN KEY (student_id) REFERENCES student(sid),
FOREIGN KEY (course_id) REFERENCES course(cid)
); INSERT INTO score(student_id, course_id, number) VALUES
(1,1,60),
(1,2,59),
(2,2,100); SELECT * FROM score; DELETE FROM score WHERE sid=6; # 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;
SELECT B.student_id FROM
(SELECT score.student_id,score.number FROM score INNER JOIN course ON score.course_id = course.cid WHERE course.cname="生物") as A
INNER JOIN
(SELECT score.student_id,score.number FROM score INNER JOIN course ON score.course_id = course.cid WHERE course.cname="物理") as B
ON A.student_id=B.student_id AND A.number>B.number; # 3、查询平均成绩大于60分的同学的学号和平均成绩;
INSERT INTO score(student_id, course_id, number) VALUES(2,3,56),
(3,1,46),(3,2,59),(3,3,71),(4,1,90),(4,2,27);
SELECT student_id,avg(number) 平均成绩 FROM score GROUP BY student_id HAVING avg(number)>60;
# 4、查询所有同学的学号、姓名、选课数、总成绩;
SELECT student.sid,student.sname,count(student.class_id) 选课数,sum(number) 总成绩
FROM student INNER JOIN score ON student.sid = score.student_id GROUP BY sname ORDER BY 总成绩; # 5、查询姓“波”的老师的个数;
SELECT count(*) 波老师个数 FROM teacher WHERE tname LIKE "波%";
INSERT INTO teacher(tname) VALUES ("波大");
# 6、查询没学过“叶平”老师课的同学的学号、姓名;
-- 得到所有同学学过的课程及其对应的老师,然后对应筛选
INSERT INTO teacher(tname) VALUES("叶平");
INSERT INTO course(cname, teacher_id) VALUES("历史",5);
INSERT INTO score(student_id, course_id, number) VALUES(1,5,81); SELECT student.sid,student.sname,A.tname FROM score
INNER JOIN student ON score.student_id = student.sid
INNER JOIN (SELECT DISTINCT course.cid,course.teacher_id,course.cname,teacher.tname FROM
course INNER JOIN teacher ON course.teacher_id = teacher.tid) AS A ON score.course_id = A.cid
GROUP BY A.tname HAVING A.tname!="叶平"; # 7、查询学过“1”并且也学过编号“2”课程的同学的学号、姓名;
INSERT INTO student(sname,gender,class_id) VALUES("张三","男",3);
INSERT INTO score(student_id, course_id, number) VALUES(5,2,63); SELECT B.sid 学号,B.sname 姓名 FROM
(SELECT student.sid,student.sname FROM score INNER JOIN student ON score.student_id = student.sid
WHERE course_id=1) AS A
INNER JOIN
(SELECT student.sid,student.sname FROM score INNER JOIN student ON score.student_id = student.sid
WHERE course_id=2) AS B ON A.sid = B.sid; # 8、查询学过“叶平”老师所教的所有课的同学的学号、姓名;
SELECT * FROM student; SELECT * FROM score;
INSERT INTO student(sname, gender, class_id) VALUES("王五","男",3);
INSERT INTO score(student_id, course_id, number) VALUES(8,5,93); SELECT student.sid 学号,student.sname 姓名 FROM score INNER JOIN student
ON score.student_id = student.sid AND course_id=5; # 9、查询课程编号“2”的成绩比课程编号“1”课程低的所有同学的学号、姓名;
INSERT INTO student(sname, gender, class_id) VALUES("赵六","女",2);
INSERT INTO score(student_id, course_id, number) VALUES(9,2,65),(9,1,70);
UPDATE score SET number = 65 WHERE sid=21; SELECT * FROM student WHERE
(SELECT number FROM score WHERE student.sid=score.student_id AND score.course_id=2)<
(SELECT number FROM score WHERE student.sid=score.student_id AND score.course_id=1); # 10、查询有课程成绩小于60分的同学的学号、姓名;
SELECT student.sid,student.sname FROM student
INNER JOIN score ON student.sid = score.student_id WHERE score.number<60 GROUP BY student.sname;
INSERT INTO student(sname, gender, class_id) VALUES("钢镚","男",2);
INSERT INTO score(student_id, course_id, number) VALUES(4,3,48);
# 11、查询没有学全所有课的同学的学号、姓名;
-- 测试
SELECT count(cid) 总课程数 FROM course;
SELECT count(course_id) study_course FROM score GROUP BY student_id;
SELECT * FROM (SELECT count(cid) 总课程数 FROM course) AS A
INNER JOIN
(SELECT count(course_id) study_course FROM score GROUP BY student_id) AS B
ON A.总课程数 = B.study_course; -- 正确答案
SELECT student.sid,student.sname,count(course_id) 学习课程数 FROM score INNER JOIN student
ON score.student_id = student.sid GROUP BY student_id
HAVING count(course_id)=(SELECT count(cid) 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;
#
# 13、查询至少学过学号为“001”同学所有一门课的其他同学学号和姓名;
#
# 14、查询和“2”号的同学学习的课程完全相同的其他同学学号和姓名; # 15、删除学习“叶平”老师课的SC表记录;
#
# 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;
#
# 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;
#
# 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(number) 最高分,min(number) 最低分 FROM score GROUP BY course_id; INSERT INTO score(student_id, course_id, number) VALUES(2,1,76);
# 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;
-- 分析求出平均成绩并计算及格率
SELECT avg(number) 课程平均成绩 FROM score GROUP BY course_id;
SELECT count(student_id) 各科不及格人数 FROM score WHERE number>60 GROUP BY course_id;
SELECT count(student_id) 各科总人数 FROM score GROUP BY course_id;
-- 答案如下
SELECT A.课程平均成绩,B.各科不及格人数/C.各科总人数 AS 及格率 FROM
(SELECT course_id,avg(number) 课程平均成绩 FROM score GROUP BY course_id) AS A
INNER JOIN
(SELECT course_id,count(student_id) 各科不及格人数 FROM score WHERE number>60 GROUP BY course_id) AS B
INNER JOIN
(SELECT course_id,count(student_id) 各科总人数 FROM score GROUP BY course_id) AS C
ON A.course_id = B.course_id AND A.course_id = C.course_id ORDER BY A.课程平均成绩; # 20、课程平均分从高到低显示(显示任课老师);
SELECT teacher.tname,course.cname FROM teacher,course WHERE course.teacher_id = teacher.tid;
SELECT avg(number) 课程平均分 FROM teacher,score GROUP BY course_id;
-- 答案如下
SELECT A.tname 任课老师,B.课程平均分 FROM
(SELECT teacher.tname,course.cname,course.cid FROM teacher,course WHERE course.teacher_id = teacher.tid) AS A
INNER JOIN
(SELECT avg(number) 课程平均分,teacher.tname,score.course_id FROM teacher,score GROUP BY course_id) AS B
ON A.cid=B.course_id ORDER BY B.课程平均分 DESC; # 21、查询各科成绩前三名的记录:(不考虑成绩并列情况)
SELECT * FROM score ORDER BY course_id; # 22、查询每门课程被选修的学生数;
-- 分析 根据成绩根据课程进行分类,按照学生ID统计次数
SELECT score.course_id 课程ID,count(student_id) 课程人次 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.sid,student.sname FROM
(SELECT * FROM score GROUP BY student_id HAVING count(student_id)=1) AS A
INNER JOIN student ON A.student_id = student.sid;
# 24、查询男生、女生的人数;
SELECT boy.男,girl.女 FROM
(SELECT count(gender) 男 FROM student WHERE gender="男") as boy,
(SELECT count(gender) 女 FROM student WHERE gender="女") as girl; # 25、查询姓“张”的学生名单;
SELECT sid,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 姓名,count(sname) 同名人数 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 A.cid 课程号,B.课程平均分 FROM
(SELECT teacher.tname,course.cname,course.cid FROM teacher,course WHERE course.teacher_id = teacher.tid) AS A
INNER JOIN
(SELECT avg(number) 课程平均分,teacher.tname,score.course_id FROM teacher,score GROUP BY course_id) AS B
ON A.cid=B.course_id ORDER BY B.课程平均分 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.sid 学号,student.sname 姓名,avg(score.number) 平均成绩 FROM
student INNER JOIN score ON student.sid = score.student_id GROUP BY sname; # 29、查询课程名称为“生物”,且分数低于60的学生姓名和分数;
SELECT course.cname 课程,student.sname 姓名,score.number 分数 FROM score INNER JOIN student INNER JOIN course
ON score.student_id = student.sid AND score.course_id=course.cid
WHERE course.cname="生物" AND score.number<60; # 30、查询课程编号为3且课程成绩在80分以上的学生的学号和姓名;
SELECT student.sid 学号,student.sname 姓名 FROM score INNER JOIN student
ON score.student_id = student.sid AND score.course_id=3 AND score.number>80; # 31、求选了课程的学生人数
INSERT INTO student(sname, gender, class_id) VALUES("李四","男",3);
INSERT INTO student(sname, gender, class_id) VALUES("胜七","女",2);
INSERT INTO score(student_id, course_id, number) VALUES(7,1,74);
-- 答案如下
SELECT count(A.student_id) 选课人数 FROM (SELECT DISTINCT student_id FROM score) AS A; # 32、查询选修“苍空”老师所授课程的学生中,成绩最高的学生姓名及其成绩;
SELECT student.sname 姓名,max(number) 成绩 FROM score INNER JOIN student INNER JOIN course INNER JOIN teacher
ON score.student_id=student.sid AND score.course_id=course.cid AND course.teacher_id=teacher.tid
WHERE teacher.tname="苍空"; # 33、查询各个课程及相应的选修人数;
SELECT score.course_id 课程ID,count(student_id) 课程人次 FROM score 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;
#
# 35、查询每门课程成绩最好的前两名;
#
# 36、检索至少选修两门课程的学生学号;
-- 分析 根据学生ID进行分组,统计课程出现次数,筛选课程次数大于等于2的学生
SELECT student_id FROM score GROUP BY student_id HAVING count(course_id)>=2;
# 37、查询全部学生都选修的课程的课程号和课程名;
SELECT score.student_id 学生ID,course.cname 所选课程,course.cid 课程ID FROM score INNER JOIN course
ON score.course_id = course.cid ORDER BY 学生ID; # 38、查询没学过“叶平”老师讲授的任一门课程的学生姓名;
#
# 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 A.student_id 学号,B.平均成绩 FROM
(SELECT student_id FROM score WHERE number<60 GROUP BY student_id HAVING count(number)>=2) AS A
INNER JOIN
(SELECT student_id,avg(number) 平均成绩 FROM score GROUP BY student_id) AS B ON A.student_id=B.student_id; # 40、检索“1”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 course_id=1 AND number<60;
# 41、删除“2”同学的“1”课程的成绩;
-- 此题已答

  

mysql考试总结的更多相关文章

  1. mysql考试复习

    基础创建 字段自动编号auto_increment ( 单词补充:increment 定期的加薪; 增量; 增加) 考点 添加自增 alter table [表名] modify [字段(id)] i ...

  2. 如何学习MySQL数据库管理员(OCP)认证(转)

    如何学习MySQL数据库管理员(OCP)认证 转自:   甲骨文专家中,MySQL 5.6数据库管理员( OCP )证明您的安装和优化MySQL服务器,设置复制和安全,执行数据库备份及性能优化和保护M ...

  3. mysql复习-来源考试

    mysql复习-   No1 .登录和权限 (一)常用命令1.登录mysqlmysql -h localhost -u root -p 2.重启mysqlservice mysql restart 延 ...

  4. 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 +-- ...

  5. MySQL和PHP基础考试错题回顾

    13.关于exit( )与die( )的说法正确的是( B) C A.当exit( )函数执行会停止执行下面的脚本,而die()无法做到 B.当die()函数执行会停止执行下面的脚本,而exit( ) ...

  6. MySQL OCP 考试,一个不错的网站

    http://www.aiotestking.com/oracle/category/exam-1z0-883-mysql-5-6-database-administrator/page/10/ 里面 ...

  7. 基于java开发jsp+ssm+mysql实现的在线考试系统 源码下载

    实现的关于在线考试的功能有:用户前台:用户注册登录.查看考试信息.进行考试.查看考试成绩.查看历史考试记录.回顾已考试卷.修改密码.修改个人信息等,后台管理功能(脚手架功能不在这里列出),科目专业管理 ...

  8. java jsp实现网络考试系统(mysql)

    java网络考试系统 功能:可进行学生.管理员登录,学生考试.管理员出卷.列表分页 @ 目录 java网络考试系统 实现效果 主要代码实现 写在最后 实现效果 主要代码实现 package cn.it ...

  9. ssm+mysql+jsp打造在线考试系统WeKnow-学生端

    一.登陆模块 前台提交账号和密码传到后台处理控制层 1.1 首先是控制器 @RequestMapping(value="/studentLogin", method=Request ...

随机推荐

  1. Rookey.Frame v1.0快速开发平台-用户登录

    上一次介绍的了Rookey.Frame v1.0快速开发平台的整体功能,接下来会对各个功能点进行解析说明,今天给大家介绍下系统登录功能. 用户登录 系统中基本上所有功能页面都是从后台代码拼接后返回的, ...

  2. mysql字符编码相关

    mysql> show variables like '%character%'; +--------------------------+--------------------------- ...

  3. java 接口请求返回通用json

    public class BaseResponseBody { // 不能添加属性 仅做泛型使用 } public class ResponseBase<T extends BaseRespon ...

  4. linux 笔记(一)

    1.Linux 安装3ython3 1.1 下载 wget https://www.python.org/ftp/python/3.6.2/Python-3.6.2.tgz 1.2 解压 tar -z ...

  5. vue引入Vue-Awesome

    Vue-Awesome建立在Font Awesome上 v4.5.0,取决于Vue.js v2.0.1 +,所以安装了Vue-Awesome就可以直接使用Font Awesome的字体图标. 1.安装 ...

  6. NBUT 1221 Intermediary

    最短路,三进制状态压缩. $dis[i][j]$表示到$i$节点,每个中介用了几次的情况下的最小花费,跑最短路即可. #include<cstdio> #include<cstrin ...

  7. Netstat -tln 命令是Linux查看端口使用情况

    Netstat -tln 命令是Linux查看端口使用情况

  8. HDU 6205[计算几何,JAVA]

    题目链接[http://acm.hdu.edu.cn/showproblem.php?pid=6206] 题意: 给出不共线的三个点,和一个点(x,y),然后判断(x,y)在不在这三个点组成的圆外. ...

  9. 【BZOJ 4527】 4527: K-D-Sequence (线段树)

    4527: K-D-Sequence Time Limit: 20 Sec  Memory Limit: 256 MBSubmit: 145  Solved: 59 Description 我们称一个 ...

  10. BZOJ1086 SCOI2005王室联邦

    想学树上莫队结果做了个树分块. 看完题后想到了菊花图的形状认为无解,结果仔细一瞧省会可以在外省尴尬 对于每一颗子树进行深搜,一旦遇到加在一起大小达到B则将它们并为一省,因为他子树搜完以后没有分出块的大 ...