script

[Former Member]

Hi Gurus,

cau you help me how to write logic for below problem

BOTTLEAMOUNT CAUFVD-MATNR CHAR 18

BUT LAST TWO DIGITS OF THE MATERIAL SHOULD BE CONVERTED TO A SINGLE DIGIT WHERE APPLICABLE AND CONCATENATED WITH A "C"

EX: 0986-03456-01

01 = 1C THEN

BOTTLE AMOUNT = 1C (IF ITS ZERO HERE )

IF ITS NOT ZERO THERE THEN WHTS THE LOGIC I HAVE WRITE FOR ABOVE 2 CONDITIONS

PLZ REVERT BACK ME

THANKS & REGARDS

DIVYA

[Former Member]

Check this sample code

data:

bottleamount like caufvd-matnr,

v_whole_length type i,

v_substring_length type i,

material_1(16),

material_2(2).

data:

begin of i_material_split occurs 0,

material(18) type c,

end of i_material_split.

move '0986-03456-01' to bottleamount.

v_whole_length = strlen( bottleamount ).

if v_whole_length gt 2.

v_substring_length = v_whole_length - 2.

material_2 = bottleamount+v_substring_length(2).

material_1 = bottleamount+0(v_substring_length).

write 😕 material_1, material_2.

endif.

*-another alternative using SPLIT

split bottleamount at '-' into table i_material_split.

loop at i_material_split.

write:/ i_material_split-material.

endloop.

*-you can then concatenate

[Former Member]

check this sample code..

data: matnr like mara-matnr,

var type i,

var1 type i.

matnr = '000000000045608'.

var = strlen( matnr ).

var1 = var - 2.

if matnr+var1(1) = 0.

var1 = var1 + 1.

concatenate matnr+var1(1) 'C' into matnr.

write:/ matnr.

else.

*- write ur logic here

endif.