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题解 | #分组过滤练习题#

select replace(SUBSTRING(start_time,1,7),"-",""), round(count(distinct uid, date_format(submit_time, '%Y%m%d')) / count(distinct uid),2), count(distin...

Mysql

2021-11-17

0 217

题解 | #分组过滤练习题#

select device_id,up1.university,up1.gpa from user_profile as up1,(select university,min(gpa)as gpa from user_profile group by university)as up2 where ...

Mysql

2021-10-20

0 191

题解 | #分组过滤练习题#

select device_id,up1.university,up1.gpa from user_profile as up1,(select university,min(gpa)as gpa from user_profile group by university)as up2 where ...

Mysql

2021-10-20

0 218

题解 | #分组过滤练习题#

select university,difficult_level,count(up.device_id)/ count(distinct(up.device_id)) from user_profile as up,question_practice_detail as qpd,question_...

Mysql

2021-10-18

0 187

题解 | #分组过滤练习题#

select university,count(question_id)/ count(distinct up.device_id) as avga from user_profile as up,question_practice_detail as qpd where up.device_id ...

Mysql

2021-10-15

0 190

题解 | #分组过滤练习题#

select qpd.device_id,question_id,result from question_practice_detail as qpd,user_profile as up where qpd.device_id=up.device_id and university="浙江大学"

Mysql

2021-10-15

0 226

题解 | #获得积分最多的人(二)#

select distinct music_name from follow as f,music_likes as ml,music as m where f.user_id='1' and f.follower_id=ml.user_id and ml.music_id=m.id and ml....

Mysql

2021-10-15

0 251

题解 | #获得积分最多的人(二)#

select user_id,name,sum from ( select user_id,name,sum(if(type='reduce',-1grade_num,grade_num))as sum, (select sum(if(type='reduce',-1grade_num,grade_...

Mysql

2021-10-15

0 193

题解 | #获得积分最多的人(二)#

select user_id,name,s from (select user_id,name,sum(grade_num) as s,m from grade_info as gi,user as u,(select sum(grade_num)as m from grade_info group...

Mysql

2021-10-15

0 211

题解 | #牛客的课程订单分析(三)#

select grade from class_grade a4 , ( select * from (select grade as g1 from (select sum(number)as sum,round(sum(number)/2) as s,if(sum(number)%2=0,rou...

Mysql

2021-10-15

0 239