Question

我有一个如下所示的数据集：

{{1}}

我想计算每个user_id从数据集中最早的day_session开始的连续日会话数。因此用户xyz789将返回1，abc123将返回14并且def456将返回38. TIA

Answer 1

根据您的数据，您可以这样做：

select t.user, count(*)
from t left join
     (select user, min(day_session) as minds
      from t
      where day_diff > 1
      group by user
     )  tt
     on t.user = tt.user
where tt.minds is null or t.day_session < tt.minds
group by t.user;

Answer 2

对于它的价值，这是一个使用SQL Server的解决方案：

declare @order table (rowid int identity, user_id varchar(max), day_session datetime, isconsecutive int)



 insert @order

 select distinct user_id, day_session, 

   case 

    when 

     lag(user_id, 1,0) over (order by user_id desc,                                     day_session)=user_id and 

     lag(day_session, 1,0) over 

  (order by user_id desc, day_session)=dateadd(dd,-1,day_session)  

    then 1 

    else 0 

   end 

from #temp 

order by user_id desc, day_session



declare @holding table (user_id varchar(max), consday int, day_session datetime)



declare @iterator int=1 

declare @userid varchar(max) 

declare @userid2 varchar(max) 

declare @isconsecutive int=1 

while @iterator<=(select max(rowid) from @order)

begin   

  select @userid=user_id, @isconsecutive=isconsecutive  

  from @order 

  where rowid=@iterator

if @isconsecutive=0 and 

(select user_id from @order where rowid=@iterator-1)=

(select user_id from @order where rowid=@iterator)

select @userid2=@userid+'_'+cast(@iterator as varchar)



if (select user_id from @order where rowid=@iterator-1)<>(select user_id from @order where rowid=@iterator)

select @userid2=null



 insert @holding

 select isnull(@userid2, @userid) ,  isconsecutive, day_session

 from @order 

 where rowid=@iterator



 set @iterator=@iterator+1

 end 



select user_id, max(cnt) maxconsdays from(

select substring(user_id, 1, len(user_id)-charindex('_', reverse(user_id)))user_id, sum(consday)+1 cnt from @holding

group by user_id)a

group by user_id

order by user_id desc

MySQL计算从最早的一天开始的连续天数

2 个答案: