topic Re: Replace in Application Development and Automation Discussions

Replace

Former Member — Wed, 25 Jul 2007 12:45:30 GMT

Hello ,

my requirement is if i have a string i want to chec if the 4th character from right is a dot or not.....if yes then replace it with comma.

if str = '123.546.236'

then it shud be 123.546,236.

also the contents of field are not of fixed length.ie. str = 12.365 or 12356.000.

Kindly provide the solution

regards

Pranali.

Re: Replace

seshatalpasai_madala — Wed, 25 Jul 2007 12:49:30 GMT

Hi,

use the Function module

STRING_REVERSE to reverse the string

and then use

REPLACE FIRST OCCURANCE OF '.' IN str WITH ','.

Again use

STRING_REVERSE

Regards,

Sesh

Re: Replace

Former Member — Wed, 25 Jul 2007 12:50:01 GMT

Use this code:


data len type i.
compute len = strlen( str ).
len = len - 4.

if str+len(1) eq '.'.
 str+len(1) = ','.
endif.

Please mark points if the solution was useful.

Regards,

Manoj

Re: Replace

Former Member — Wed, 25 Jul 2007 12:50:13 GMT

DATA : LV_CHAR TYPE C,

STR(10) TYPE C.

STR = 'Hel.oo'.

LV_CHAR = STR+3(1).

IF LV_CHAR EQ '.' .

STR+3(1) = ','.

ENDIF.

WRITE STR.

Thanks,

Mahesh

Re: Replace

Former Member — Wed, 25 Jul 2007 12:51:50 GMT

Hey Pranali,

Try this out,

Get the lenght of your string str

len = strlen( str ).

Less 4, to get the right position

len = len - 4.

*Using offset get one caracter at that position, if it is a dot, change to comma

if str+len(1) = '.'.

str+len(1) = ','.

endif.

Hope this helps.

Regards,

Marcelo Moreira

Re: Replace

RichHeilman — Wed, 25 Jul 2007 12:51:57 GMT

data: str(20) type c.
data: len type i.


str = '123.546.236'.

len = ( strlen( str ) - 4 ).

if str+len(1) = '.'.
  str+len(1) = ','.
endif.

write:/ str.

Regards,

Rich Heilman

Re: Replace

kesavadas_thekkillath — Wed, 25 Jul 2007 12:52:15 GMT

if str cs '.'.

REPLACE '.' WITH ',' INTO str.

endif.

reward if usefull.....

Re: Replace

Former Member — Wed, 25 Jul 2007 12:54:37 GMT

REPORT ztest.

DATA : v_str TYPE string VALUE '123.546.236',
       v_char(250),
         v_len TYPE i.

v_char = v_str.
v_len = STRLEN( v_char ).

v_len = v_len - 4.

IF v_char+v_len(1) EQ '.'.
  v_char+v_len(1) = ','.
ENDIF.

v_str = v_char.

WRITE : v_str.

Re: Replace

Former Member — Wed, 25 Jul 2007 12:54:39 GMT

v_len = stelen (str).

v_len = v_len - 3.

if str+0(v_len) = '.'.

replace str+0(v_len) with ','.

endif.

Re: Replace

Former Member — Wed, 25 Jul 2007 12:55:06 GMT

If str+3(1) = '.'.         " If dot
   str+3(1) = ','.                 " replace with comma
endif.

reward if useful

Regards

Prax