Extracting File

Question

Hello Guru,
We are going to extract data from excel(csv) file.File will be everyday - deliveries_YYYYMMDD.csv.File will be delivered everyday in one directory.My question is how can we extract this file everyday ? .We can give path to the directory but everyday file will be with different date ??? What should we do ?
Please advice me.Thanks in advance.

Answer 1

In the routine of Datasource you will set as following:
orm compute_flat_file_filename
  using    p_infopackage  type rslogdpid
           p_datasource   type rsoltpsourcer
           p_logsys       type rsslogsys
  changing p_filename     type RSFILENM
           p_subrc        like sy-subrc.
$$ begin of routine - insert your code only below this line        -
This routine will be called by the adapter,
when the infopackage is executed.
  DATA Z_FILENAME TYPE RSFILENM.
  DATA ZFILE_13(100).
  DATA: ZDATE LIKE SY-DATUM,        DD(2),
        MM(2),
        YY(4).
  ZDATE = SY-DATUM.
  Z_FILENAME =
  your path of local client.
  YY = ZDATE+0(4).
  MM = ZDATE+4(2).
  DD = ZDATE+6(2).
  CONCATENATE YY MM DD INTO ZFILE_13.
  CONCATENATE Z_FILENAME  ZFILE_13  '.csv' INTO Z_FILENAME. 
p_filename = Z_FILENAME.
   p_subrc = 0.
$$ end of routine - insert your code only before this line         -
endform.
Create an infopackage with similar setting as that of datasource.
Edited by: gang qin on Feb 1, 2011 10:48 PM