Question

我有一个餐桌订单，上面有关于下订单的时间以及下订单的人的信息。

   order_timestamp     user_id  
 -------------------- --------- 
  1-JAN-20 02.56.12        123  
  3-JAN-20 12.01.01        533  
  23-JAN-20 08.42.18       123  
  12-JAN-20 02.53.59       238  
  19-JAN-20 02.33.72        34

我想使用此信息每天计算在过去31天中只下过一次订单的不重复用户数，得出的表为

    date      distinct_user_count  
 ----------- --------------------- 
  1-JAN-20                      8   
  2-JAN-20                     10  
  3-JAN-20                     11  
(i.e in the 31 days before and including 1st jan 2020, 8 unique users ordered only once, etc...)

简单地说，每天-31，计算该时段内每个用户的订单数（表中的条目），如果该计数仅为1，则将该用户计为初始开始日期。

我可以编写查询来将仅订购一次的人计数为：

with temp as (
  select 
    user_id,
    count(*) as order_count
  from 
    orders 
  where 
    trunc(order_timestamp) >= trunc(systimestamp - interval '31' day)
  group by 
    user_id
)
select 
  user_id, 
  order_count 
from 
  temp 
 where 
  login_count=1

，但不确定如何实施每个日期的计数。请您协助我完成/撰写查询吗？感谢您的预先支持。

Answer 1

您可以使用grouping和self join这两个级别，如下所示：

Select dt, count(1) as cnt
  from
(Select distinct trunc(t1.order_timestamp) as dt, 
        t1.user_id
   From your_table t1
   join your_table t2
     On t1.user_id = t2.user_id
    And trunc(t2.order_timestamp) between trunc(t1.order_timestamp - interval '31' day) 
                                 and trunc(t1.order_timestamp)
 Group by t1.user_id, trunc(t1.order_timestamp)
 Having count(1) = 1)
Group by dt;

或者您可以按以下方式使用NOT EXISTS：

Select trunc(t1.order_timestamp),
       Count(1) as cnt
  From your_table t1
 Where not exists 
  (Select 1
     From your_table t2
     Where t1.rowid <> t2.row_id
       And t1.user_id = t2.user_id 
       And trunc(t2.order_timestamp) between trunc(t1.order_timestamp - interval '31' day) 
                                  and trunc(t1.order_timestamp)
Group by trunc(t1.order_timestamp)

每天根据过去31天的状况滚动唯一计数

1 个答案: