Splitting a string into multiple parts from particular position (string has no delimiter )

[Former Member]

Hello Gurus,

I need to make string of having length 1000 into multiple parts from a particular position which has some text without any delimiter and i want to populate each part into an internal table which has one filed...

eg : string = 'sap abap sap abap sap abap sap abap sap abap sap abap sap abap sap sap sap abap sap abap'

suppose i want to start splitting from 40th position...assume that i have 3 parts

and these 3 parts i have put into an internal table..

lt_itab = [sap abap sap abap sap abap sap abap,

              sap abap sap abap sap abap sap abap,

               sap abap sap sap sap abap sap abap]

plese help!

any help would be Appreciated ...thank you

[Former Member]

Hi Nagendra,

I'm just sharing code which need to changed accordling. Just an idea

DATA : w_str TYPE string VALUE 'abapaaaaaaaaaaaaaaaaaaaaaaaaa aabaaaaaaaaaaaaaaaaaaqqqqq',

           part  TYPE n VALUE 3,

          w_lengh(4) TYPE n,

          offset(3) TYPE n,

          start(3)  TYPE n,

         wa_area(100)   TYPE c .

* String lengh

w_lengh = strlen( w_str ).

offset = w_lengh  DIV part .

start = 0.

DO part times.

   wa_area = w_str+start(offset).

* APPEND wa_rea to t_table.

   start = start + offset.

ENDDO.

Thanks & Regards,

Arun

[Former Member]

Thank you so much for your reply

[mayur_priyan]

Hi,

Try this code.

TYPES: BEGIN OF ty,

        b TYPE string,

        END OF ty,

        tt TYPE STANDARD TABLE OF ty.

DATA: a TYPE string VALUE 'aaa bbb ccc ddd eee fff ggg hhh iii jjj kkk lll',

       tb TYPE tt,

       wa TYPE ty,

       off TYPE i,

       len TYPE i,

       count TYPE i,

       count1 TYPE i,

       temp TYPE i.

len = 8.                      "Position

off = 0.

count = strlen( a ).

count1 = count / 8.

temp = len - 1.

DO count1 times.

  temp = off + len.

  IF temp > count.

   wa-b = a+off.

  ELSE.

   wa-b = a+off(len).

  ENDIF.

  off = off + len.

  SHIFT wa-b LEFT DELETING LEADING space.

  APPEND wa to tb.

ENDDO.

LOOP AT  tb INTO wa.

write: / wa-b.

ENDLOOP.

[dibyajeeban_jena]

hi,

check below coding and use  fm - SWA_STRING_SPLIT ,,

* Split each line of it_longstrings into lines with length 10

* The resuklting lines are palced in table it_string_components

LOOP AT it_longstrings.

  CLEAR it_string_components. REFRESH it_string_components.

  CALL FUNCTION 'SWA_STRING_SPLIT'

    EXPORTING

      input_string                       = it_longstrings-line

      max_component_length               = 10

*     TERMINATING_SEPARATORS             =

*     OPENING_SEPARATORS                 =

    TABLES

      string_components                  = it_string_components

*   EXCEPTIONS

*     MAX_COMPONENT_LENGTH_INVALID       = 1

*     OTHERS                             = 2

            .

  IF sy-subrc <> 0.

* MESSAGE ID SY-MSGID TYPE SY-MSGTY NUMBER SY-MSGNO

*         WITH SY-MSGV1 SY-MSGV2 SY-MSGV3 SY-MSGV4.

  ENDIF.

* Add the lines to table

  LOOP AT it_string_components.

    it_shortstrings-line = it_string_components-str.

    APPEND it_shortstrings.

  ENDLOOP.

ENDLOOP.

END-OF-SELECTION.

  LOOP AT  it_shortstrings.

    WRITE: / it_shortstrings-line.

  ENDLOOP.

Regards DJ

[Former Member]

Thank you...

[Former Member]

Here's one more way to do it.

1. DATA str TYPE string VALUE 'first012345679801234567890123456789'.
2. DATA lt TYPE TABLE OF char10.
3. DO.
4.   TRY .
5.       APPEND str(10) TO lt.
6.       str = str+10.
7.     CATCH cx_sy_range_out_of_bounds.
8.       APPEND str TO lt.
9.       EXIT.
10.   ENDTRY.
11. ENDDO.

/.

[mayur_priyan]

I wonder why you have marked your 'Thank you' replies as Correct/Helpful Answers.

[rosenberg_eitan]

Hi,

Some time when I try to solve a problem I wake up the next day with the solution.

My theory is that I have "A SPLIT personality" and the other guy is much smarter then I am.....

Regards.

[matt]

Fixed

[mayur_priyan]

Well that was a quick response. Thanks