从 Oracle 中的每个给定当前日期滚动 7 个日历日期

Question

给定：

calendar_date Date_Key  RowN 
03-OCT-21     20211003  1
03-OCT-21     20211003  2
03-OCT-21     20211003  3
04-OCT-21     20211004  4
07-OCT-21     20211007  5
07-OCT-21     20211007  6
09-OCT-21     20211009  7
10-OCT-21     20211010  8
13-OCT-21     20211013  9
13-OCT-21     20211013  10
13-OCT-21     20211013  11
15-OCT-21     20211015  12
17-OCT-21     20211017  13

需要：从每个给定的当前日期开始滚动 7 个日历日期

这是我的代码，没有给我正确的结果

select z.calendar_date,z.date_key,ceil(z.rown/7) as Grp from
(select b.calendar_date,b.date_key ,rownum as RowN
from Samp.cal_date_dim a, Samp.cal_date_dim b
where b.date_key-a.date_key>0 and b.date_key-a.date_key<=6)z

选码说明：如果是03 OCT开始，那么03 OCT - 09 OCT -> Group 1。如果是04 OCT开始，那么04 OCT - 10 OCT -> Group 2。

OutPut 要求或期望的结果

calendar_date  Grp  
03-OCT-21      1    
03-OCT-21      1    
03-OCT-21      1    
04-OCT-21      1    
07-OCT-21      1  
07-OCT-21      1  
09-OCT-21      1

04-OCT-21      2    
07-OCT-21      2  
07-OCT-21      2  
09-OCT-21      2
10-OCT-21      2

07-OCT-21      3
07-OCT-21      3
09-OCT-21      3
10-OCT-21      3
13-OCT-21      3
13-OCT-21      3
13-OCT-21      3

Answer 1

您似乎在要求自加入，以及 dense_rank():

select dd.*, cd.seqnum as grp
from (select calendar_date, Date_Key,
             dense_rank() over (order by calendar_date) as seqnum
      from Samp.cal_date_dim
      group by calendar_date, Date_Key
     ) cd join
     Samp.cal_date_dim dd
     on dd.calendar_date >= cd.calendar_date and
        dd.calendar_date < cd.calendar_date + interval '7' day;

从 Oracle 中的每个给定当前日期滚动 7 个日历日期

Rolling over 7 Calendar date from each given current date in Oracle

sql

oracle

date

count

rolling-computation