topic Re: Replace last 4 digits in Application Development and Automation Discussions

Replace last 4 digits

Former Member — Thu, 03 Jan 2008 15:07:40 GMT

Hi Guys

Wish you all a happy new year

I have a requirement where I have to replace the last 4 digits of a material number by XXXX. The material number can be of any lenght not more than 18.

Does any one know how to codefor this situation?

<REMOVED BY MODERATOR>

Thanks

Edited by: Sameer Ahmed on Jan 3, 2008 4:10 PM

Edited by: Alvaro Tejada Galindo on Jan 3, 2008 10:10 AM

Re: Replace last 4 digits

former_member156446 — Thu, 03 Jan 2008 15:12:29 GMT

lv_len = strln(matnr).

lv_off = lv_len - 4.

matnr = matnr+lv_off

concatenate matnr 'XXXX' into matnr.

Re: Replace last 4 digits

Former Member — Thu, 03 Jan 2008 15:13:43 GMT


DATA: LONG TYPE I.
LONG = STRLEN( MATERIAL ).
LONG = LONG - 4.
MATERIAL+LONG(4) = 'XXXX'.

Greetings,

Blag.

Re: Replace last 4 digits

Former Member — Thu, 03 Jan 2008 15:15:11 GMT

try this...

DATA : matnr LIKE marc-matnr.

DATA : mat_len TYPE i.

matnr = 'ABCDSFYU1313'.

mat_len = STRLEN( matnr ).

mat_len = mat_len - 4.

matnr+mat_len(4) = 'XXXX'.

WRITE 😕 matnr.

Re: Replace last 4 digits

Former Member — Thu, 03 Jan 2008 15:17:18 GMT

Hi,

data : lv_len

shift matnr left deleting leading '0'.

lv_len = STRLEN( MATNR ).

lv_len = lv_len - 4.

MATNR+lv_len(4) = 'XXXX'

Lokesh

Edited by: Lokesh Aggarwal on Jan 3, 2008 3:17 PM

Re: Replace last 4 digits

Former Member — Thu, 03 Jan 2008 20:49:05 GMT

Thanks Guys.