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题解 | #满足条件的用户的试卷完成数和题目练习数#

select a.uid, count(distinct case when year(b.submit_time) = '2021' then b.id else null end) as exam_cnt, count(distinct ca...

Mysql

2022-04-27

0 358

题解 | #2021年11月每天的人均浏览文章时长#

select date_format(in_time,'%Y-%m-%d') as dt, round(sum(timestampdiff(second, in_time, out_time)) / count(distinct uid), 1) as avg_viiew_len_sec fr...

Mysql

2022-04-26

1 344

题解 | #平均活跃天数和月活人数#

select author, date_format(start_time,'%Y-%m') month, round(sum(case when if_follow=1 then 1 when if_follow=2 then -1 else 0 end)...

Mysql

2022-04-25

0 278

题解 | #平均活跃天数和月活人数#

select date_format(submit_time, '%Y%m') as month, round( count(distinct uid, date_format(submit_time, '%Y%m%d')) / count(distinct uid), 2) as avg_ac...

Mysql

2022-04-19

0 327

题解 | #修改表#

alter table user_info add school varchar(15) after level; alter table user_info change job profession varchar(10); alter table user_info modify achiev...

Mysql

2022-04-18

0 219

题解 | #计算用户的平均次日留存率#

select count(distinct device_id) as did_cnt, count(question_id) as question_cnt from question_practice_detail where year(date)=2021 and month(date)...

Mysql

2022-04-06

0 228

题解 | #计算用户的平均次日留存率#

select qd.difficult_level as difficult_level, sum(case when qpd.result = "right" then 1 else 0 end)/ count(qpd.device_id) as correct_rate from us...

Mysql

2022-04-06

0 243

题解 | #计算男生人数以及平均GPA#

select count(gender) as male_num, round (avg(gpa), 1) as avg_gpa from user_profile where gender = "male" 注意缩进，count函数与括号之间不要留空格

Mysql

2022-04-05

0 185