CentOS Linux release 7.4.1708 (Core) CentOS Linux release 7.4.1708 (Core) [root@mysql ~]# mysql --version mysql Ver 8.0.25 for Linux on x86_64 (MySQL Community Server - GPL)
select st.*,sc.s_score as'语文' ,sc2.s_score '数学' from student st leftjoin score sc on sc.s_id=st.s_id and sc.c_id='01' leftjoin 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
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select st.*,sc.s_score '语文',sc2.s_score '数学'from student st leftjoin score sc on sc.s_id=st.s_id and sc.c_id='01' leftjoin score sc2 on sc2.s_id=st.s_id and sc2.c_id='02' where sc.s_score<sc2.s_score;
– 3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩
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select st.s_id,st.s_name,ROUND(AVG(sc.s_score),2) cjScore from student st leftjoin score sc on sc.s_id=st.s_id groupby st.s_id havingAVG(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)
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select st.s_id,st.s_name,(casewhen ROUND(AVG(sc.s_score),2) isnullthen0else ROUND(AVG(sc.s_score)) end ) cjScore from student st leftjoin score sc on sc.s_id=st.s_id groupby st.s_id havingAVG(sc.s_score)<60orAVG(sc.s_score) isNULL;
– 5、查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩
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select st.s_id,st.s_name,count(c.c_id),( casewhenSUM(sc.s_score) isnullorsum(sc.s_score)="" then0elseSUM(sc.s_score) end) from student st leftjoin score sc on sc.s_id =st.s_id leftjoin course c on c.c_id=sc.c_id groupby st.s_id;
– 6、查询”李”姓老师的数量
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select t.t_name,count(t.t_id) from teacher t groupby t.t_id having t.t_name like "李%";
– 7、查询学过”张三”老师授课的同学的信息
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select st.*from student st leftjoin score sc on sc.s_id=st.s_id leftjoin course c on c.c_id=sc.c_id leftjoin teacher t on t.t_id=c.t_id where t.t_name="张三";
– 8、查询没学过”张三”老师授课的同学的信息
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-- 张三老师教的课 select c.*from course c leftjoin teacher t on t.t_id=c.t_id where t.t_name="张三";
-- 有张三老师课成绩的st.s_id select sc.s_id from score sc where sc.c_id in (select c.c_id from course c leftjoin teacher t on t.t_id=c.t_id where t.t_name="张三");
-- 不在上面查到的st.s_id的学生信息,即没学过张三老师授课的同学信息 select st.*from student st where st.s_id notin( select sc.s_id from score sc where sc.c_id in (select c.c_id from course c leftjoin teacher t on t.t_id=c.t_id where t.t_name="张三") );
– 9、查询学过编号为”01”并且也学过编号为”02”的课程的同学的信息
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select st.*from student st innerjoin score sc on sc.s_id = st.s_id innerjoin course c on c.c_id=sc.c_id and c.c_id="01" where st.s_id in ( select st2.s_id from student st2 innerjoin score sc2 on sc2.s_id = st2.s_id innerjoin course c2 on c2.c_id=sc2.c_id and c2.c_id="02" )
-- 网友提供的思路(厉害呦~): SELECT st.* FROM student st INNERJOIN score sc ON sc.`s_id`=st.`s_id` GROUPBY st.`s_id` HAVINGSUM(IF(sc.`c_id`="01" OR sc.`c_id`="02" ,1,0))>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
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select st.*from student st innerjoin score sc on sc.s_id = st.s_id innerjoin course c on c.c_id=sc.c_id and c.c_id="01" where st.s_id notin ( select st2.s_id from student st2 innerjoin score sc2 on sc2.s_id = st2.s_id innerjoin course c2 on c2.c_id=sc2.c_id and c2.c_id="02" )
– 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 notin ( select st.s_id from student st innerjoin 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 innerjoin 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 innerjoin 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 leftjoin Score S on st.s_id = S.s_id groupby st.s_id havingcount(c_id)<(selectcount(c_id) from Course)
– 12、查询至少有一门课与学号为”01”的同学所学相同的同学的信息
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selectdistinct st.*from student st leftjoin score sc on sc.s_id=st.s_id where sc.c_id in ( select sc2.c_id from student st2 leftjoin score sc2 on sc2.s_id=st2.s_id where st2.s_id ='01' )
– 13、查询和”01”号的同学学习的课程完全相同的其他同学的信息
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select st.*from student st leftjoin score sc on sc.s_id=st.s_id groupby st.s_id having group_concat(sc.c_id) = ( select group_concat(sc2.c_id) from student st2 leftjoin score sc2 on sc2.s_id=st2.s_id where st2.s_id ='01' )
– 14、查询没学过”张三”老师讲授的任一门课程的学生姓名
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select st.s_name from student st where st.s_id notin ( select sc.s_id from score sc innerjoin course c on c.c_id=sc.c_id innerjoin teacher t on t.t_id=c.t_id and t.t_name="张三" )
– 15、查询两门及其以上不及格课程的同学的学号,姓名及其平均成绩
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select st.s_id,st.s_name,avg(sc.s_score) from student st leftjoin score sc on sc.s_id=st.s_id where sc.s_id in ( select sc.s_id from score sc where sc.s_score<60or sc.s_score isNULL groupby sc.s_id havingCOUNT(sc.s_id)>=2 ) groupby 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
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select st.*,sc.s_score from student st innerjoin score sc on sc.s_id=st.s_id and sc.c_id="01" and sc.s_score<60 orderby sc.s_score desc
– 17、按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩
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– 可加round,casewhenthenelseend 使显示更完美
select st.s_id,st.s_name,avg(sc4.s_score) "平均分",sc.s_score "语文",sc2.s_score "数学",sc3.s_score "英语" from student st leftjoin score sc on sc.s_id=st.s_id and sc.c_id="01" leftjoin score sc2 on sc2.s_id=st.s_id and sc2.c_id="02" leftjoin score sc3 on sc3.s_id=st.s_id and sc3.c_id="03" leftjoin score sc4 on sc4.s_id=st.s_id groupby st.s_id orderbySUM(sc4.s_score) desc
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" ,((selectcount(s_id) from score where s_score>=60and c_id=c.c_id )/(selectcount(s_id) from score where c_id=c.c_id)) "及格率" ,((selectcount(s_id) from score where s_score>=70and s_score<80and c_id=c.c_id )/(selectcount(s_id) from score where c_id=c.c_id)) "中等率" ,((selectcount(s_id) from score where s_score>=80and s_score<90and c_id=c.c_id )/(selectcount(s_id) from score where c_id=c.c_id)) "优良率" ,((selectcount(s_id) from score where s_score>=90and c_id=c.c_id )/(selectcount(s_id) from score where c_id=c.c_id)) "优秀率" from course c leftjoin score sc on sc.c_id=c.c_id leftjoin score sc2 on sc2.c_id=c.c_id leftjoin score sc3 on sc3.c_id=c.c_id groupby 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
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select c1.s_id,c1.c_id,c1.c_name,@score:=c1.s_score,@i:=@i+1from (select c.c_name,sc.*from course c leftjoin score sc on sc.c_id=c.c_id where c.c_id="01" orderby sc.s_score desc) c1 , (select@i:=0) a unionall select c2.s_id,c2.c_id,c2.c_name,c2.s_score,@ii:=@ii+1from (select c.c_name,sc.*from course c leftjoin score sc on sc.c_id=c.c_id where c.c_id="02" orderby sc.s_score desc) c2 , (select@ii:=0) aa unionall select c3.s_id,c3.c_id,c3.c_name,c3.s_score,@iii:=@iii+1from (select c.c_name,sc.*from course c leftjoin score sc on sc.c_id=c.c_id where c.c_id="03" orderby sc.s_score desc) c3; set@iii=0;
– 20、查询学生的总成绩并进行排名
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select st.s_id,st.s_name ,(casewhensum(sc.s_score) isnullthen0elsesum(sc.s_score) end) from student st leftjoin score sc on sc.s_id=st.s_id groupby st.s_id orderbysum(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
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select t.t_id,t.t_name,c.c_name,avg(sc.s_score) from teacher t leftjoin course c on c.t_id=t.t_id leftjoin score sc on sc.c_id =c.c_id groupby t.t_id orderbyavg(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
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select a.*from ( select st.*,c.c_id,c.c_name,sc.s_score from student st leftjoin score sc on sc.s_id=st.s_id innerjoin course c on c.c_id =sc.c_id and c.c_id="01" orderby sc.s_score desc LIMIT 1,2 ) a unionall select b.*from ( select st.*,c.c_id,c.c_name,sc.s_score from student st leftjoin score sc on sc.s_id=st.s_id innerjoin course c on c.c_id =sc.c_id and c.c_id="02" orderby sc.s_score desc LIMIT 1,2) b unionall select c.*from ( select st.*,c.c_id,c.c_name,sc.s_score from student st leftjoin score sc on sc.s_id=st.s_id innerjoin course c on c.c_id =sc.c_id and c.c_id="03" orderby sc.s_score desc LIMIT 1,2) c
select c.c_id,c.c_name ,((selectcount(1) from score sc where sc.c_id=c.c_id and sc.s_score<=100and sc.s_score>80)/(selectcount(1) from score sc where sc.c_id=c.c_id )) "100-85" ,((selectcount(1) from score sc where sc.c_id=c.c_id and sc.s_score<=85and sc.s_score>70)/(selectcount(1) from score sc where sc.c_id=c.c_id )) "85-70" ,((selectcount(1) from score sc where sc.c_id=c.c_id and sc.s_score<=70and sc.s_score>60)/(selectcount(1) from score sc where sc.c_id=c.c_id )) "70-60" ,((selectcount(1) from score sc where sc.c_id=c.c_id and sc.s_score<=60and sc.s_score>=0)/(selectcount(1) from score sc where sc.c_id=c.c_id )) "60-0" from course c orderby 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<60 inner join course c on c.c_id=sc.c_id and c.c_name =”数学” – 35、查询所有学生的课程及分数情况;
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 left join course c on c.c_id =sc.c_id order by st.s_id,c.c_name – 36、查询任何一门课程成绩在70分以上的姓名、课程名称和分数
select st2.s_id,st2.s_name,c2.c_name,sc2.s_score 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 st2.s_id in( select st.s_id from student st left join score sc on sc.s_id=st.s_id group by st.s_id having min(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<60 inner join course c on c.c_id=sc.c_id – 38、查询课程编号为01且课程成绩在80分以上的学生的学号和姓名
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.c_id=”01” 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)<=2 order by a.c_id – 43、统计每门课程的学生选修人数(超过5人的课程才统计)。要求输出课程号和选修人数,查询结果按人数降序排列, – 若人数相同,按课程号升序排列
select sc.c_id,count(1) from score sc left join course c on c.c_id=sc.c_id group by c.c_id having count(1)>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
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’))