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REPLACE вставляет пробел



 
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stn
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Joined: 18 Jan 2008
Posts: 31
PostPosted: Thu Oct 02, 2008 11:12 am    Post subject: REPLACE вставляет пробел Reply with quote
Добрый день.
Формирую файл для банк-клиента по заданному шаблону. Если для замены передается пустое полу, то REPLACE вставляет одинарный пробел. Нужно, чтобы ничего не вставлял.

Code:
DATA: ls(255), out(1024).
CONSTANTS: c_template LIKE out VALUE
    '&,&,&,@&@,@&@,@&@,@&@,&,&,@&@,@&@,' &
    '&,@&@,@&@,@&@,@&@,@&@,@&@,@&@,&,' &
    '&,&,&,&,@&@,@&@,@&@,@&@,&,&,&,&,' &
    '&,&,&,@&@,@&@,@&@,@&@,@&@,@&@,@&@'.
   
out = c_template.

*1 Date 10 Дата последней операции по счету 18/10/2006
  CLEAR ls.
  PERFORM transfer USING ls.

*2 Date 10 Дата операции. 19/10/2006
  WRITE g_date USING EDIT MASK '__/__/____'.
  PERFORM transfer USING ls.

*&---------------------------------------------------------------------*
*&      Form  transfer
*&---------------------------------------------------------------------*
*       text
*----------------------------------------------------------------------*
*      -->P_LS  text
*----------------------------------------------------------------------*
FORM transfer USING p_ls.

  DATA: len TYPE I.
  len = STRLEN( p_ls ).

  IF len EQ 0.
    REPLACE '&' LENGTH 1 WITH '' INTO outt-str.
  ELSE.
    REPLACE '&' LENGTH 1 WITH p_ls(len) INTO outt-str.
  ENDIF.
ENDFORM.                    " transfer
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vga
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Age: 165
Joined: 04 Oct 2007
Posts: 1218
Location: Санкт-Петербург
PostPosted: Fri Oct 03, 2008 3:34 pm    Post subject: Reply with quote
Иcпользовать стринг вместо char ''

Code:
FORM transfer USING p_ls.

  DATA: len TYPE I,
        s TYPE STRING.
  len = STRLEN( p_ls ).

  IF len EQ 0.
    REPLACE '&' LENGTH 1 WITH s INTO outt-str.
  ELSE.
    REPLACE '&' LENGTH 1 WITH p_ls(len) INTO outt-str.
  ENDIF.
ENDFORM.                    " transfer
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