Get directory path. Just the directory path, without the file name.

Question

Hi,
I need to export a file from a table in SAP to a directory in the PC. So i need to get a directory path.
I've been reading the threads but all the answers refer to functions F4_FILENAME, WS_FILENAME_GET, KD_GET_FILENAME_ON_F4 or to the method CL_GUI_FRONTEND_SERVICES=>FILE_SAVE_DIALOG.
But all of them require to select a file to close the window.
What I need is to get the path to the directory, not to a file.
For example:
I just need to get:
C:\TEMP
and not:
C:\TEMP\file.doc
Please help.
Thanks.

Answer 1

Pablo, use cl_gui_frontend_services=>directory_browse.
Refer this code:
PARAMETERS:
  p_path         TYPE rlgrap-filename OBLIGATORY.
DATA:
  gv_path_ini    TYPE string,
  gv_path_sel    TYPE string.
AT SELECTION-SCREEN ON VALUE-REQUEST FOR p_path.
  CALL METHOD cl_gui_frontend_services=>directory_browse
    EXPORTING
      initial_folder  = gv_path_ini
    CHANGING
      selected_folder = gv_path_sel
    EXCEPTIONS
      cntl_error      = 1
      error_no_gui    = 2
      OTHERS          = 3
  IF sy-subrc = 0.
    CALL METHOD cl_gui_cfw=>flush( ).
    IF NOT gv_path_sel IS INITIAL.
      gv_path_ini = p_path = gv_path_sel.
    ENDIF.
  ENDIF.
As a side note, always avoid any WS_* function modules as they're not Unicode compliant.
Regards