How to get filename from selected directory

Question

Helle experts,
I've a string variable from a directory with following content:
G:\SAP_R3_MRSS\SAP_MRS\FAVORIT.PDF
now I just want to select the filename, in this case FAVORIT.TXT
How can I do this ? The directory/filename may vary.
Some ideas ?
Thanks
Gerd

Answer 1

Eric's split is probably the best way, but if you need a way to break the path into drive, path, and file, you could use a standard, well-used regular expression:
PROGRAM zzfile.
DATA:
   g_regex     TYPE string VALUE '\b((?#drive)[a-z]):\\((?#folder)[^/:*?"<>|\r\n]*\\)?((?#file)[^\\/:*?"<>|\r\n]*)',
   g_path_file TYPE string VALUE 'G:\SAP_R3_MRSS\SAP_MRS\FAVORIT.PDF',
   g_drive     TYPE string,
   g_path      TYPE string,
   g_file      TYPE string.
FIND FIRST OCCURRENCE OF REGEX g_regex IN g_path_file SUBMATCHES g_drive g_path g_file IGNORING CASE.
IF sy-subrc = 0.
*  WRITE g_drive.
*  WRITE g_path.
  WRITE g_file.
ENDIF.