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【学号】连接两个临时表:
学号 物理成绩 生物成绩 然后再进行筛选 select A.student_id,sw,ty from (select student_id,num as sw from score left join course on score.course_id = course.cid where course.cname = '生物') as A left join (select student_id,num as ty from score left join course on score.course_id = course.cid where course.cname = '体育') as B on A.student_id = B.student_id where sw > if(isnull(ty),0,ty); 3、查询平均成绩大于60分的同学的学号和平均成绩;
思路:
根据学生分组,使用avg获取平均值,通过having对avg进行筛选 select student_id,avg(num) from score group by student_id having avg(num) > 60 4、查询所有同学的学号、姓名、选课数、总成绩; select score.student_id,sum(score.num),count(score.student_id),student.sname
from
score left join student on score.student_id = student.sid
group by score.student_id 5、查询姓“李”的老师的个数;
select count(tid) from teacher where tname like '李%' select count(1) from (select tid from teacher where tname like '李%') as B 6、查询没学过“叶平”老师课的同学的学号、姓名;
思路:
先查到“李平老师”老师教的所有课ID
获取选过课的所有学生ID
学生表中筛选
select * from student where sid not in (
select DISTINCT student_id from score where score.course_id in (
select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '李平老师'
)
) 7、查询学过“001”并且也学过编号“002”课程的同学的学号、姓名;
思路:
先查到既选择001又选择002课程的所有同学
根据学生进行分组,如果学生数量等于2表示,两门均已选择 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 8、查询学过“叶平”老师所教的所有课的同学的学号、姓名; 同上,只不过将001和002变成 in (叶平老师的所有课) 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;
同第1题 10、查询有课程成绩小于60分的同学的学号、姓名; select sid,sname from student where sid in (
select distinct student_id from score where num < 60
) 11、查询没有学全所有课的同学的学号、姓名;
思路:
在分数表中根据学生进行分组,获取每一个学生选课数量
如果数量 == 总课程数量,表示已经选择了所有课程 select student_id,sname
from score left join student on score.student_id = student.sid
group by student_id HAVING count(course_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;
思路:
获取 001 同学选择的所有课程
获取课程在其中的所有人以及所有课程
根据学生筛选,获取所有学生信息
再与学生表连接,获取姓名 select student_id,sname, count(course_id)
from score left join student on score.student_id = student.sid
where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id 13、查询至少学过学号为“001”同学所有课的其他同学学号和姓名;
先找到和001的学过的所有人
然后个数 = 001所有学科 ==》 其他人可能选择的更多 select student_id,sname, count(course_id)
from score left join student on score.student_id = student.sid
where student_id != 1 and course_id in (select course_id from score where student_id = 1) group by student_id having count(course_id) = (select count(course_id) from score where student_id = 1) 14、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名; 个数相同
002学过的也学过 select student_id,sname from score left join student on score.student_id = student.sid where student_id in (
select student_id from score where student_id != 1 group by student_id HAVING count(course_id) = (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(course_id) = (select count(1) from score where student_id = 1) 15、删除学习“叶平”老师课的score表记录; delete from score where course_id in (
select cid from course left join teacher on course.teacher_id = teacher.tid where teacher.name = '叶平'
) 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;
思路:
由于insert 支持
inset into tb1(xx,xx) select x1,x2 from tb2;
所有,获取所有没上过002课的所有人,获取002的平均成绩 insert into score(student_id, course_id, num) select sid,2,(select avg(num) from score where course_id = 2)
from student where sid not in (
select student_id 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;
select sc.student_id,
(select num from score left join course on score.course_id = course.cid where course.cname = "生物" and score.student_id=sc.student_id) as sy,
(select num from score left join course on score.course_id = course.cid where course.cname = "物理" and score.student_id=sc.student_id) as wl,
(select num from score left join course on score.course_id = course.cid where course.cname = "体育" and score.student_id=sc.student_id) as ty,
count(sc.course_id),
avg(sc.num)
from score as sc
group by student_id 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) as max_num, min(num) as 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;
思路:case when .. then
select course_id, avg(num) as avgnum,sum(case when score.num > 60 then 1 else 0 END)/count(1)*100 as percent from score group by course_id order by avgnum asc,percent desc; 20、课程平均分从高到低显示(现实任课老师); select avg(if(isnull(score.num),0,score.num)),teacher.tname from course
left join score on course.cid = score.course_id
left join teacher on course.teacher_id = teacher.tid group by score.course_id 21、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.second_num from score left join
(
select
sid,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0,1) as first_num,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 3,1) as second_num
from
score as s1
) as T
on score.sid =T.sid
where score.num <= T.first_num and score.num >= T.second_num 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.sid, student.sname, count(1) from score left join student on score.student_id = student.sid group by course_id having count(1) = 1 24、查询男生、女生的人数;
select * from
(select count(1) as man from student where gender='男') as A ,
(select count(1) as feman from student where gender='女') as B 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) as count 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 course_id,avg(if(isnull(num), 0 ,num)) as avg from score group by course_id order by avg 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; 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.sname,score.num from score
left join course on score.course_id = course.cid
left join student on score.student_id = student.sid
where score.num < 60 and course.cname = '生物' 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 * from score where score.student_id = 3 and score.num > 80 31、求选了课程的学生人数 select count(distinct student_id) from score select count(c) from (
select count(student_id) as c 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 student on score.student_id = student.sid
where score.course_id in (select course.cid from course 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 score.course_id = course.cid
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 DISTINCT s1.course_id,s2.course_id,s1.num,s2.num from score as s1, score as s2 where s1.num = s2.num and s1.course_id != s2.course_id; 35、查询每门课程成绩最好的前两名; select score.sid,score.course_id,score.num,T.first_num,T.second_num from score left join
(
select
sid,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 0,1) as first_num,
(select num from score as s2 where s2.course_id = s1.course_id order by num desc limit 1,1) as second_num
from
score as s1
) as T
on score.sid =T.sid
where score.num <= T.first_num and score.num >= T.second_num 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(student_id) > 1 37、查询全部学生都选修的课程的课程号和课程名;
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;
select student_id,student.sname from score
left join student on score.student_id = student.sid
where score.course_id not in (
select cid from course left join teacher on course.teacher_id = teacher.tid where tname = '张磊老师'
)
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) 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 course_id = 1 and student_id = 2
MySQL练习题参考答案的更多相关文章
- 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 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 及 答案
--1.学生表 Student(S,Sname,Sage,Ssex) --S 学生编号,Sname 学生姓名,Sage 出生年月,Ssex 学生性别 --2.课程表 Course(C,Cname,T ...
- MySQL 练习题 附答案,未完
综合练习题 表结构 整合一下方便查看 teacher student course scors 练习题 1.自行创建测试数据 create table student( sid int prima ...
- MySQL练习题及答案
一.现有三张数据库表,分别为部门表.员工表.部门和员工关系表 1.部门表CREATE TABLE `t_dept` ( `id` int(8) NOT NULL AUTO_INCREMENT, `de ...
- 数据库---MySQL练习题及答案
一. 设有一数据库,包括四个表:学生表(Student).课程表(Course).成绩表(Score)以及教师信息表(Teacher).四个表的结构分别如表1-1的表(一)~表( ...
- 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习题和答案
MySQL测试题 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,再把符合 ...
随机推荐
- js正则表达式校验非负浮点数:^[1-9]\d*\.\d*|0\.\d*[1-9]\d*|0?\.0+|0$
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...
- C#开发微信门户及应用(19)-微信企业号的消息发送(文本、图片、文件、语音、视频、图文消息等)
我们知道,企业号主要是面向企业需求而生的,因此内部消息的交流显得非常重要,而且发送.回复消息数量应该很可观,对于大企业尤其如此,因此可以结合企业号实现内部消息的交流.企业号具有关注安全.消息无限制等特 ...
- 是时候 UWP 了 !
什么是 UWP? 很多程序员都有一个梦想:希望自己开发的软件能够轻而易举的在所有平台上运行,而不是把同样的需求,用不同的技术.工具重新开发才能够运行在所有平台上.这就是跨平台,很多软件从业者都在为这个 ...
- openresty 前端开发入门一
OpenResty ™ 是一个基于 Nginx 与 Lua 的高性能 Web 平台,其内部集成了大量精良的 Lua 库.第三方模块以及大多数的依赖项.用于方便地搭建能够处理超高并发.扩展性极高的动态 ...
- 生成Tab键或逗号分隔的CSV
<?php header("Content-type:text/csv;charset=utf-8"); header("Content-Disposition:a ...
- java分解质因数
package test; import java.util.Scanner; public class Test19 { /** * 分析:对n进行分解质因数,应先找到一个最小的质数k * 最小 ...
- mysql awr 1.0.5 GA正式版发布
1.0.5变更内容 1.修复centos 7下swap值不正确:2.中文乱码:3.begin/end snap下拉显示Mysql启动时间:4.两次快照间不能重启过:5.新增tab页面查看mysql存储 ...
- win7 由ie8升级ie11时安装不成功的一个原因
纯净win7系统更新好补丁好 升级ie一直提示不成功,之前重来没有遇到过.官方提示的解决办法: https://support.microsoft.com/zh-cn/kb/2872074 根据内容初 ...
- arcgis api for js入门开发系列三地图工具栏(含源代码)
上一篇实现了demo的地图加载展示,在上篇实现的基础上,新增了地图工具栏以及通用地图控件功能,比如地图框选缩放.地图漫游.清空.量算工具.地图导航控件.地图比例尺控件.地图鹰眼图等等,总共分为5个部分 ...
- Atitit.日志系统slf4j的使用
Atitit.日志系统slf4j的使用 SLF4J: Class path contains multiple SLF4J bindings. SLF4J: Found binding in [jar ...