mysql50道题查询和,Mysql Sql 语句练习题（50道）

-- 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 stleft 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 stleft 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 stleft join score sc on sc.s_id=st.s_idgroup by st.s_id having AVG(sc.s_score)<60 or AVG(sc.s_score) is NULL-- 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select st.s_id,st.s_name,count(c.c_id),( case when SUM(sc.s_score) is null or sum(sc.s_score)='' then 0 else SUM(sc.s_score) end) from student stleft join score sc on sc.s_id =st.s_id left join course c on c.c_id=sc.c_idgroup by st.s_id-- 6、查询'李'姓老师的数量 select t.t_name,count(t.t_id) from teacher tgroup by 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 select st.* from student st left join score sc on sc.s_id=st.s_idleft join course c on c.c_id=sc.c_idleft join 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 -- 张三老师教的课 select c.* from course c left join 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 left join 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 not in( select sc.s_id from score sc where sc.c_id in (select c.c_id from course c left join 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select st.* from student st inner join score sc on sc.s_id= st.s_idinner join 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 inner join score sc2 on sc2.s_id = st2.s_idinner join 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 stINNER JOIN score sc ON sc.`s_id`=st.`s_id`GROUP BY st.`s_id`HAVING SUM(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select st.* from student st inner join score sc on sc.s_id= st.s_idinner 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_idinner 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 stleft join Score Son st.s_id= S.s_idgroup by st.s_idhaving 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<60order by 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 -- 可加round,case when then else end 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 stleft 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'left join score sc3 on sc3.s_id=st.s_id and sc3.c_id='03'left join score sc4 on sc4.s_id=st.s_idgroup by st.s_id order by SUM(sc4.s_score) desc-- 18.查询各科成绩最高分、最低分和平均分：以如下形式显示：课程ID，课程name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率-- 及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90select 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 c_id=c.c_id )/(select count(s_id) from score where c_id=c.c_id)) '中等率',((select count(s_id) from score where 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)) '优良率',((select count(s_id) from score where 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 cleft 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_idwhere c.c_id='01' order by sc.s_score desc) c1 ,(select @i:=0) aunion 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_idwhere c.c_id='02' order by sc.s_score desc) c2 ,(select @ii:=0) aa union allselect 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_idwhere 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 stleft join score sc on sc.s_id=st.s_idgroup 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_idgroup by t.t_idorder 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 stleft join score sc on sc.s_id=st.s_idinner join course c on c.c_id =sc.c_id and c.c_id='01'order by sc.s_score desc LIMIT 1,2 ) aunion allselect b.* from (select st.*,c.c_id,c.c_name,sc.s_score from student stleft join score sc on sc.s_id=st.s_idinner join course c on c.c_id =sc.c_id and c.c_id='02'order by sc.s_score desc LIMIT 1,2) bunion allselect c.* from (select st.*,c.c_id,c.c_name,sc.s_score from student stleft join score sc on sc.s_id=st.s_idinner 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_score<=100 and sc.s_score>80)/(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_score<=85 and sc.s_score>70)/(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_score<=70 and sc.s_score>60)/(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<=60 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 stleft join score sc on sc.s_id=st.s_idgroup 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) aunion 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) bunion allselect 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_idinner join student st on st.s_id=c.c_idgroup 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_idinner 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 cinner 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 stleft join score sc on sc.s_id=st.s_idgroup 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 stinner join score sc on sc.s_id=st.s_id and sc.s_score<60inner 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 stleft join score sc on sc.s_id=st.s_idleft join course c on c.c_id =sc.c_idorder 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 st2left join score sc2 on sc2.s_id=st2.s_idleft 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 stinner join score sc on sc.s_id=st.s_id and sc.s_score<60inner 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 stinner 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 cinner join score sc on sc.c_id=c.c_idgroup 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 stinner join score sc on sc.s_id=st.s_idinner 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 desclimit 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_idleft join course c on c.c_id=sc.c_idwhere (select count(1) from student st2 left join score sc2 on sc2.s_id=st2.s_idleft join course c2 on c2.c_id=sc2.c_idwhere 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 stleft join score sc on sc.s_id=st.s_idinner join course c on c.c_id=sc.c_id and c.c_id='01'order by sc.s_score desc limit 0,2) aunion allselect b.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student stleft join score sc on sc.s_id=st.s_idinner join course c on c.c_id=sc.c_id and c.c_id='02'order by sc.s_score desc limit 0,2) bunion allselect c.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student stleft join score sc on sc.s_id=st.s_idinner 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 scleft join course c on c.c_id=sc.c_idgroup by c.c_id having count(1)>5order 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_idgroup 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_idgroup 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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mysql50道题查询和,Mysql Sql 语句练习题（50道）