Mysql Sql 语句练习题(50道)
来源:脚本之家
时间:2023-01-07 12:06:16 223浏览 收藏
怎么入门数据库编程?需要学习哪些知识点?这是新手们刚接触编程时常见的问题;下面golang学习网就来给大家整理分享一些知识点,希望能够给初学者一些帮助。本篇文章就来介绍《Mysql Sql 语句练习题(50道)》,涉及到Mysql练习题,有需要的可以收藏一下
表名和字段
–1.学生表
Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别
–2.课程表
Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号
–3.教师表
Teacher(t_id,t_name) –教师编号,教师姓名
–4.成绩表
Score(s_id,c_id,s_score) –学生编号,课程编号,分数
测试数据
--建表 --学生表 CREATE TABLE `Student`( `s_id` VARCHAR(20), `s_name` VARCHAR(20) NOT NULL DEFAULT '', `s_birth` VARCHAR(20) NOT NULL DEFAULT '', `s_sex` VARCHAR(10) NOT NULL DEFAULT '', PRIMARY KEY(`s_id`) ); --课程表 CREATE TABLE `Course`( `c_id` VARCHAR(20), `c_name` VARCHAR(20) NOT NULL DEFAULT '', `t_id` VARCHAR(20) NOT NULL, PRIMARY KEY(`c_id`) ); --教师表 CREATE TABLE `Teacher`( `t_id` VARCHAR(20), `t_name` VARCHAR(20) NOT NULL DEFAULT '', PRIMARY KEY(`t_id`) ); --成绩表 CREATE TABLE `Score`( `s_id` VARCHAR(20), `c_id` VARCHAR(20), `s_score` INT(3), PRIMARY KEY(`s_id`,`c_id`) ); --插入学生表测试数据 insert into Student values('01' , '赵雷' , '1990-01-01' , '男'); insert into Student values('02' , '钱电' , '1990-12-21' , '男'); insert into Student values('03' , '孙风' , '1990-05-20' , '男'); insert into Student values('04' , '李云' , '1990-08-06' , '男'); insert into Student values('05' , '周梅' , '1991-12-01' , '女'); insert into Student values('06' , '吴兰' , '1992-03-01' , '女'); insert into Student values('07' , '郑竹' , '1989-07-01' , '女'); insert into Student values('08' , '王菊' , '1990-01-20' , '女'); --课程表测试数据 insert into Course values('01' , '语文' , '02'); insert into Course values('02' , '数学' , '01'); insert into Course values('03' , '英语' , '03'); --教师表测试数据 insert into Teacher values('01' , '张三'); insert into Teacher values('02' , '李四'); insert into Teacher values('03' , '王五'); --成绩表测试数据 insert into Score values('01' , '01' , 80); insert into Score values('01' , '02' , 90); insert into Score values('01' , '03' , 99); insert into Score values('02' , '01' , 70); insert into Score values('02' , '02' , 60); insert into Score values('02' , '03' , 80); insert into Score values('03' , '01' , 80); insert into Score values('03' , '02' , 80); insert into Score values('03' , '03' , 80); insert into Score values('04' , '01' , 50); insert into Score values('04' , '02' , 30); insert into Score values('04' , '03' , 20); insert into Score values('05' , '01' , 76); insert into Score values('05' , '02' , 87); insert into Score values('06' , '01' , 31); insert into Score values('06' , '03' , 34); insert into Score values('07' , '02' , 89); insert into Score values('07' , '03' , 98);
表数据如下
student 学生表:
s_id | s_name | s_birth | s_sex |
---|---|---|---|
01 | 赵雷 | 1990-01-01 | 男 |
02 | 钱电 | 1990-12-21 | 男 |
03 | 孙凤 | 1990-05-20 | 男 |
04 | 李云 | 1990-08-06 | 男 |
05 | 周梅 | 1991-12-12 | 女 |
06 | 吴兰 | 2017-12-13 | 女 |
07 | 郑竹 | 1989-07-01 | 女 |
08 | 王菊 | 1990-01-20 | 女 |
09 | 赵雷 | 1990-01-21 | 女 |
10 | 赵雷 | 1990-01-22 | 男 |
score 分数表:
s_id | c_id | s_score |
---|---|---|
01 | 01 | 80 |
01 | 02 | 90 |
01 | 03 | 99 |
02 | 01 | 70 |
02 | 02 | 60 |
02 | 03 | 80 |
03 | 01 | 80 |
03 | 02 | 80 |
03 | 03 | 80 |
04 | 01 | 50 |
04 | 02 | 30 |
04 | 03 | 20 |
05 | 01 | 76 |
05 | 03 | 87 |
06 | 01 | 31 |
06 | 03 | 34 |
07 | 03 | 89 |
07 | 01 | 98 |
course 课程表
c_id | c_name | t_id |
---|---|---|
01 | 语文 | 02 |
02 | 数学 | 01 |
03 | 英语 | 03 |
teacher 老师表:
t_id | t_name |
---|---|
01 | 张三 |
02 | 李四 |
03 | 王五 |
-- 准备条件,去掉 sql_mode 的 ONLY_FULL_GROUP_BY 否则此种情况下会报错: -- Expression #1 of select list is not in group by clause and contains nonaggregated column 'userinfo. -- 原因: -- MySQL 5.7.5和up实现了对功能依赖的检测。如果启用了only_full_group_by SQL模式(在默认情况下是这样), -- 那么MySQL就会拒绝选择列表、条件或顺序列表引用的查询,这些查询将引用组中未命名的非聚合列,而不是在功能上依赖于它们。 -- (在5.7.5之前,MySQL没有检测到功能依赖项,only_full_group_by在默认情况下是不启用的。关于前5.7.5行为的描述,请参阅MySQL 5.6参考手册。) -- 执行以下个命令,可以查看 sql_mode 的内容。 SHOW SESSION VARIABLES; SHOW GLOBAL VARIABLES; select @@sql_mode; -- 更改 set global sql_mode='STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION'; set session sql_mode='STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION';
练习题和sql
-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数 select st.*,sc.s_score as '语文' ,sc2.s_score '数学' from student st left join score sc on sc.s_id=st.s_id and sc.c_id='01' left join score sc2 on sc2.s_id=st.s_id and sc2.c_id='02' where sc.s_score>sc2.s_score -- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数 select st.*,sc.s_score '语文',sc2.s_score '数学' from student st left join score sc on sc.s_id=st.s_id and sc.c_id='01' left join score sc2 on sc2.s_id=st.s_id and sc2.c_id='02' where sc.s_score=60 -- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩 -- (包括有成绩的和无成绩的) select st.s_id,st.s_name,(case when ROUND(AVG(sc.s_score),2) is null then 0 else ROUND(AVG(sc.s_score)) end ) cjScore from student st left join score sc on sc.s_id=st.s_id group by st.s_id having AVG(sc.s_score)1 -- 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息 select st.* from student st inner join score sc on sc.s_id = st.s_id inner join course c on c.c_id=sc.c_id and c.c_id="01" where st.s_id not in ( select st2.s_id from student st2 inner join score sc2 on sc2.s_id = st2.s_id inner join course c2 on c2.c_id=sc2.c_id and c2.c_id="02" ) -- 11、查询没有学全所有课程的同学的信息 -- 太复杂,下次换一种思路,看有没有简单点方法 -- 此处思路为查学全所有课程的学生id,再内联取反面 select * from student where s_id not in ( select st.s_id from student st inner join score sc on sc.s_id = st.s_id and sc.c_id="01" where st.s_id in ( select st2.s_id from student st2 inner join score sc2 on sc2.s_id = st2.s_id and sc2.c_id="02" ) and st.s_id in ( select st2.s_id from student st2 inner join score sc2 on sc2.s_id = st2.s_id and sc2.c_id="03" )) -- 来自一楼网友的思路,左连接,根据学生id分组过滤掉 数量小于 课程表中总课程数量的结果(show me his code),简洁不少。 select st.* from Student st left join Score S on st.s_id = S.s_id group by st.s_id having count(c_id)=2 ) group by st.s_id -- 16、检索"01"课程分数小于60,按分数降序排列的学生信息 select st.*,sc.s_score from student st inner join score sc on sc.s_id=st.s_id and sc.c_id="01" and sc.s_score=60,中等为:70-80,优良为:80-90,优秀为:>=90 select c.c_id,c.c_name,max(sc.s_score) "最高分",MIN(sc2.s_score) "最低分",avg(sc3.s_score) "平均分" ,((select count(s_id) from score where s_score>=60 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) "及格率" ,((select count(s_id) from score where s_score>=70 and s_score=80 and s_score=90 and c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) "优秀率" from course c left join score sc on sc.c_id=c.c_id left join score sc2 on sc2.c_id=c.c_id left join score sc3 on sc3.c_id=c.c_id group by c.c_id -- 19、按各科成绩进行排序,并显示排名(实现不完全) -- mysql没有rank函数 -- 加@score是为了防止用union all 后打乱了顺序 select c1.s_id,c1.c_id,c1.c_name,@score:=c1.s_score,@i:=@i+1 from (select c.c_name,sc.* from course c left join score sc on sc.c_id=c.c_id where c.c_id="01" order by sc.s_score desc) c1 , (select @i:=0) a union all select c2.s_id,c2.c_id,c2.c_name,c2.s_score,@ii:=@ii+1 from (select c.c_name,sc.* from course c left join score sc on sc.c_id=c.c_id where c.c_id="02" order by sc.s_score desc) c2 , (select @ii:=0) aa union all select c3.s_id,c3.c_id,c3.c_name,c3.s_score,@iii:=@iii+1 from (select c.c_name,sc.* from course c left join score sc on sc.c_id=c.c_id where c.c_id="03" order by sc.s_score desc) c3; set @iii=0; -- 20、查询学生的总成绩并进行排名 select st.s_id,st.s_name ,(case when sum(sc.s_score) is null then 0 else sum(sc.s_score) end) from student st left join score sc on sc.s_id=st.s_id group by st.s_id order by sum(sc.s_score) desc -- 21、查询不同老师所教不同课程平均分从高到低显示 select t.t_id,t.t_name,c.c_name,avg(sc.s_score) from teacher t left join course c on c.t_id=t.t_id left join score sc on sc.c_id =c.c_id group by t.t_id order by avg(sc.s_score) desc -- 22、查询所有课程的成绩第2名到第3名的学生信息及该课程成绩 select a.* from ( select st.*,c.c_id,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id =sc.c_id and c.c_id="01" order by sc.s_score desc LIMIT 1,2 ) a union all select b.* from ( select st.*,c.c_id,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id =sc.c_id and c.c_id="02" order by sc.s_score desc LIMIT 1,2) b union all select c.* from ( select st.*,c.c_id,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id =sc.c_id and c.c_id="03" order by sc.s_score desc LIMIT 1,2) c -- 23、统计各科成绩各分数段人数:课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比 select c.c_id,c.c_name ,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score80)/(select count(1) from score sc where sc.c_id=c.c_id )) "100-85" ,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score70)/(select count(1) from score sc where sc.c_id=c.c_id )) "85-70" ,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score60)/(select count(1) from score sc where sc.c_id=c.c_id )) "70-60" ,((select count(1) from score sc where sc.c_id=c.c_id and sc.s_score=0)/(select count(1) from score sc where sc.c_id=c.c_id )) "60-0" from course c order by c.c_id -- 24、查询学生平均成绩及其名次 set @i=0; select a.*,@i:=@i+1 from ( select st.s_id,st.s_name,round((case when avg(sc.s_score) is null then 0 else avg(sc.s_score) end),2) "平均分" from student st left join score sc on sc.s_id=st.s_id group by st.s_id order by sc.s_score desc) a -- 25、查询各科成绩前三名的记录 select a.* from ( select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id=sc.c_id and c.c_id='01' order by sc.s_score desc LIMIT 0,3) a union all select b.* from ( select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id=sc.c_id and c.c_id='02' order by sc.s_score desc LIMIT 0,3) b union all select c.* from ( select st.s_id,st.s_name,c.c_id,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id=sc.c_id and c.c_id='03' order by sc.s_score desc LIMIT 0,3) c -- 26、查询每门课程被选修的学生数 select c.c_id,c.c_name,count(1) from course c left join score sc on sc.c_id=c.c_id inner join student st on st.s_id=c.c_id group by st.s_id -- 27、查询出只有两门课程的全部学生的学号和姓名 select st.s_id,st.s_name from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id=sc.c_id group by st.s_id having count(1)=2 -- 28、查询男生、女生人数 select st.s_sex,count(1) from student st group by st.s_sex -- 29、查询名字中含有"风"字的学生信息 select st.* from student st where st.s_name like "%风%"; -- 30、查询同名同性学生名单,并统计同名人数 select st.*,count(1) from student st group by st.s_name,st.s_sex having count(1)>1 -- 31、查询1990年出生的学生名单 select st.* from student st where st.s_birth like "1990%"; -- 32、查询每门课程的平均成绩,结果按平均成绩降序排列,平均成绩相同时,按课程编号升序排列 select c.c_id,c.c_name,avg(sc.s_score) from course c inner join score sc on sc.c_id=c.c_id group by c.c_id order by avg(sc.s_score) desc,c.c_id asc -- 33、查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩 select st.s_id,st.s_name,avg(sc.s_score) from student st left join score sc on sc.s_id=st.s_id group by st.s_id having avg(sc.s_score)>=85 -- 34、查询课程名称为"数学",且分数低于60的学生姓名和分数 select st.s_id,st.s_name,sc.s_score from student st inner join score sc on sc.s_id=st.s_id and sc.s_score=70) order by s_id -- 37、查询不及格的课程 select st.s_id,c.c_name,st.s_name,sc.s_score from student st inner join score sc on sc.s_id=st.s_id and sc.s_score=80 -- 39、求每门课程的学生人数 select c.c_id,c.c_name,count(1) from course c inner join score sc on sc.c_id=c.c_id group by c.c_id -- 40、查询选修"张三"老师所授课程的学生中,成绩最高的学生信息及其成绩 select st.*,c.c_name,sc.s_score,t.t_name from student st inner join score sc on sc.s_id=st.s_id inner join course c on c.c_id=sc.c_id inner join teacher t on t.t_id=c.t_id and t.t_name="张三" order by sc.s_score desc limit 0,1 -- 41、查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩 select st.s_id,st.s_name,sc.c_id,sc.s_score from student st left join score sc on sc.s_id=st.s_id left join course c on c.c_id=sc.c_id where ( select count(1) from student st2 left join score sc2 on sc2.s_id=st2.s_id left join course c2 on c2.c_id=sc2.c_id where sc.s_score=sc2.s_score and c.c_id!=c2.c_id )>1 -- 42、查询每门功成绩最好的前两名 select a.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id=sc.c_id and c.c_id="01" order by sc.s_score desc limit 0,2) a union all select b.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id=sc.c_id and c.c_id="02" order by sc.s_score desc limit 0,2) b union all select c.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st left join score sc on sc.s_id=st.s_id inner join course c on c.c_id=sc.c_id and c.c_id="03" order by sc.s_score desc limit 0,2) c -- 借鉴(更准确,漂亮): select a.s_id,a.c_id,a.s_score from score a where (select COUNT(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)5 order by count(1) desc,sc.c_id asc -- 44、检索至少选修两门课程的学生学号 select st.s_id from student st left join score sc on sc.s_id=st.s_id group by st.s_id having count(1)>=2 -- 45、查询选修了全部课程的学生信息 select st.* from student st left join score sc on sc.s_id=st.s_id group by st.s_id having count(1)=(select count(1) from course) -- 46、查询各学生的年龄 select st.*,timestampdiff(year,st.s_birth,now()) from student st -- 47、查询本周过生日的学生 -- 此处可能有问题,week函数取的为当前年的第几周,2017-12-12是第50周而2018-12-12是第49周,可以取月份,day,星期几(%w), -- 再判断本周是否会持续到下一个月进行判断,太麻烦,不会写 select st.* from student st where week(now())=week(date_format(st.s_birth,'%Y%m%d')) -- 48、查询下周过生日的学生 select st.* from student st where week(now())+1=week(date_format(st.s_birth,'%Y%m%d')) -- 49、查询本月过生日的学生 select st.* from student st where month(now())=month(date_format(st.s_birth,'%Y%m%d')) -- 50、查询下月过生日的学生 -- 注意:当 当前月为12时,用month(now())+1为13而不是1,可用timestampadd()函数或mod取模 select st.* from student st where month(timestampadd(month,1,now()))=month(date_format(st.s_birth,'%Y%m%d')) -- 或 select st.* from student st where (month(now()) + 1) mod 12 = month(date_format(st.s_birth,'%Y%m%d'))
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评论列表
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- 爱笑的汉堡
- 很棒,一直没懂这个问题,但其实工作中常常有遇到...不过今天到这,看完之后很有帮助,总算是懂了,感谢作者大大分享技术文章!
- 2023-01-31 23:46:00
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- 愤怒的樱桃
- 这篇文章内容真是及时雨啊,太细致了,赞 👍👍,收藏了,关注楼主了!希望楼主能多写数据库相关的文章。
- 2023-01-30 22:14:14
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- 有魅力的树叶
- 这篇文章内容真是及时雨啊,太全面了,很有用,收藏了,关注老哥了!希望老哥能多写数据库相关的文章。
- 2023-01-25 11:30:27
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- 生动的钢笔
- 很有用,一直没懂这个问题,但其实工作中常常有遇到...不过今天到这,帮助很大,总算是懂了,感谢师傅分享技术文章!
- 2023-01-24 04:29:46
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- 潇洒的酒窝
- 这篇文章内容太及时了,太细致了,很好,已加入收藏夹了,关注作者了!希望作者能多写数据库相关的文章。
- 2023-01-21 11:34:19
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- 妩媚的月亮
- 好细啊,码起来,感谢up主的这篇技术贴,我会继续支持!
- 2023-01-11 13:09:13
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- 俊逸的枕头
- 太细致了,收藏了,感谢博主的这篇文章内容,我会继续支持!
- 2023-01-09 19:53:44
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- 无奈的御姐
- 太给力了,一直没懂这个问题,但其实工作中常常有遇到...不过今天到这,帮助很大,总算是懂了,感谢作者大大分享技术贴!
- 2023-01-09 14:40:51