Concatenating Zero to string

[Former Member]

Hi,

I need to know how to add zero to a string.

I have VAR1(18) TYPE C in which the value is 47896.24 now I need to have out put as 000000000047896.24

I tried the FM 'CONVERSION_EXIT_ALPHA_INPUT' which is working for numeric only and my data type is C.

Thanks in Advance.

Regards,

Kiran

[seshatalpasai_madala]

Hi,

Execute this code and see

DATA: str1 type String,

str2 type String.

DATA: char_num(8) type c value '47896.24'.

SPLIT char_num AT '.' INTO str1 str2.

DATA: numc(15) type n.

numc = str1.

concatenate numc '.' str2 into str1.

write: str1.

Regards,

Sesh

[Former Member]

hi,

you can use CONCATENATE for that.

CONCATENATE '00000000000' 47896.24 into var1.

if it helps REWARD...

rgds

harris

[Former Member]

Hi kiran,

do one thing split at '.' and move the values to variables

use conversion routine to numeric value

and again use concatenate.

Please close thread if it is helpful,

Best Regds,

kiran.M

[former_member189059]

Hello Kiran,

try this, it should solve your problem

data: lv_oldvar type string.
lv_oldvar = '47896.24'.

data: lv_total type i value 18. " total length required
data: lv_diff type i.
data: lv_newvar type string.

lv_diff = lv_total - strlen( lv_oldvar ).

lv_newvar = lv_oldvar.

do lv_diff times.
  concatenate '0' lv_newvar into lv_newvar.
enddo.

write:/ lv_newvar.

[Former Member]

Hi kiran,

You can do it like following:

DATA: VAR1(18) TYPE C VALUE '47896.24'.

DATA: VAR2(15) TYPE C,

VAR3(3) TYPE C.

SPLIT VAR1 AT '.' INTO VAR2 VAR3.

CALL FUNCTION 'CONVERSION_EXIT_ALPHA_INPUT'

EXPORTING

INPUT = VAR2

IMPORTING

OUTPUT = VAR2.

CONCATENATE VAR2 '.' VAR3 INTO VAR1.

WRITE VAR1.

Thanks and regards,

Dharitree

[Former Member]

HI,

Try this,it will work fine.

<b>DATA : VAR1(18) TYPE C VALUE '47896.24'.

TRANSLATE VAR1 USING ' 0'.

WRITE : VAR1.</b>

Output:

<b>000000000047896.24</b>

Regards,

Padmam.

[Former Member]

Hi kiran,

try this:

data: null(18) value '000000000000000000'.

data: v1(18) value '47896.24'.

data: i type i.

data: j type I.

i = strlen( v1 ).

j = 18 - i.

null+j = v1(i).

regards, Dieter

[Former Member]

Thanks every one

[Former Member]

Hi Kiran

where is my points

[former_member189059]

Hello Kiran C,

Did my method work ?