高级资深攻城狮
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题解 | #考试分数(二)#
select g.id,g.job,g.score from grade g, ( select job ,round(sum(score)*1.0 / count(score),3) as s from grade&n...
Mysql
2021-09-29
0
194
题解 | #牛客每个人最近的登录日期(六)#
select u.name,date,(select sum(number)from passing_number pn1 where pn1.date<=pn.date and pn1.user_id=pn.user_id) from passing_number as pn,user u ...
Mysql
2021-09-29
0
208
题解 | #牛客每个人最近的登录日期(四)#
select r1.date,ifnull(r.id, 0) from( select date from login group by date)as r1left join( select l.date,count( l.user_id)as id from login l ...
Mysql
2021-09-29
0
166
题解 | #牛客每个人最近的登录日期(四)#
select r1.date,ifnull(r.id, 0) from( select date from login group by date)as r1left join( select l.date,if(count( l.user_id)>0,count( l.user_...
Mysql
2021-09-29
0
206
题解 | #牛客每个人最近的登录日期(一)#
select round( ( select count(*) from login where (user_id,date) in( select user_id,date_add(min(date),interval 1 day) from ...
Mysql
2021-09-29
0
206
题解 | #牛客每个人最近的登录日期(一)#
先按照 userID进行分组,查出分组中最新的登录日期,通过userID和最新的登录日期,找出其他关联信息。select u.name,c.name,l.datefrom login l,client c,user u,( select user_id,max(date)as date ...
Mysql
2021-09-29
0
153
题解 | #异常的邮件概率#
Mysql的写法: select date,round(( select count() from ( select date,type from email e where e.send_id in ( se...
Mysql
2021-09-29
0
266
题解 | #找到每个人的任务#
总之解决了问题,代码有点长:select date,( select count() from ( select date,type from email e where e.send_id in ( sele...
Sqlite
2021-09-29
0
265
题解 | #刷题通过的题目排名#
select r.id ,r.number,(select count(distinct number )from passing_number where passing_number.number>=r.number )from( select id,number from p...
Mysql
2021-09-28
0
210
题解 | #出现三次以上相同积分的情况#
select r.number from( select number ,count(number)as c from grade group by number)as rwhere r.c>=3题目需要的是积分出现的次数,那就以积分分组,统计积分出现的次数,再将超过三次的积分查...
Mysql
2021-09-28
0
239
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