replace continous numbers with space

[Former Member]

hi...

an urgent requirement...

i have an internal table, with rows containing characters of 70 length.

if i find continous 16 digits of numbers, i have to replace the numbers with space.

How am i supposed to do this?

The first row can be continued by next row record also.

So, how to check the contnuinty in the next record also.

for example...

Line 1--- thrirj1234

line 2 --- 45678thyeu

so, it has c first line and next line to is to be checked to c if it is starting wth numbers.

please do help me solve this issue..

Thank u

[matt]

Credit card numbers?

Untested - but this could work:

DATA: rec(70),
      t_tab LIKE STANDARD TABLE OF rec,
      l_checkrec(85).

FIELD-SYMBOLS: <l_rec>      LIKE rec,
               <l_next_rec> LIKE rec.

DATA: l_index      TYPE i,
      l_checklen   TYPE i,
      l_pos        TYPE i,
      l_char       TYPE c LENGTH 1,
      l_num_length TYPE i,
      l_start      TYPE i,
      l_len        TYPE i.

LOOP AT t_tab ASSIGNING <l_rec>.
  l_index = sy-tabix + 1.
  READ TABLE t_tab ASSIGNING <l_next_rec> INDEX l_index.
  IF <l_next_rec> IS ASSIGNED.
    CONCATENATE <l_rec> <l_next_rec> INTO l_checkrec.
  ELSE.
    l_checkrec = <l_rec>.
  ENDIF.
  l_checklen = STRLEN( l_checkrec ).

  DO l_checklen TIMES.                                    
    l_char = l_checkrec+l_pos(1).
    IF l_char CA '0123456789'.
      ADD 1 TO l_num_length.
    ELSE.
      CLEAR l_num_length.
    ENDIF.
    IF l_num_length EQ 16.
      l_start = l_pos - 16.
      IF l_pos LE 70.

        CLEAR <l_rec>+l_start(16).
      ELSE.
        CLEAR <l_rec>+l_start.
        l_len = l_pos - 70.
        CLEAR <l_next_rec>(l_len).
      ENDIF.
    ENDIF.
  ENDDO.
ENDLOOP.

As I say, untested - so you'll probably have to play with some of the numbers to get it exactly right, but I think the principle is obvious.

matt