Question

select retention
 , Sum(count(retention)) over(order by retention desc) as Num_Users_Retained
  from (select player_id
         , round(init_dtime-create_dtime,0) as retention
      from player
     where Trunc(Create_Dtime) >= To_Date('2012-jan-01','yyyy-mon-dd')
       and init_dtime is not null)
  Group by retention
  order by retention

此查询显示两列：第一列是今年的第一天（第0天是jan-1，第259天是今天的sep-16），第二列是保留。第0天有428,000，因为所有用户都被保留，第1天有300K（因为128K人在第一天之后没有登录），随着时间的推移，这个数字不断减少。我想为每天的第0天百分比显示另一列。这是否足够清楚，还是应该提供更多详细信息？

Answer 1

select retention
     , Num_Users_Retained
     , round(Num_Users_Retained/max(Num_Users_Retained) over() * 100, 2) as perc
  from (select retention
             , Sum(count(retention)) over(order by retention desc) as Num_Users_Retained
         from (select player_id
                    , round(init_dtime-create_dtime,0) as retention
                from player
               where Trunc(Create_Dtime) >= To_Date('2012-jan-01','yyyy-mon-dd')
                 and init_dtime is not null
              )
        Group by retention
        order by retention)
 order by retention

SQL - 以百分比显示结果

1 个答案: