Zeroes after Decimal

[Former Member]

Hi,

I am having a value as '6009.12300 ML' belongs to type C.

Now i don't want the zeroes after decimal such as 6009.123 ML .

how cud i do this?

i tried search ,split,shift,trailing zeroes but could nt get it?

Thanks,

Ponraj.s.

[Former Member]

Hello,

data: wa_input  type i,      
      wa_output type i.
 
wa_input = '1.000'.
 
CALL FUNCTION 'ROUND'  
  EXPORTING    
    INPUT              = wa_input 
  IMPORTING   
    OUTPUT             = wa_output
  EXCEPTIONS
   INPUT_INVALID       = 1
   OVERFLOW            = 2
   TYPE_INVALID        = 3
   OTHERS              = 4.
 
IF sy-subrc <> 0.
* MESSAGE ID SY-MSGID TYPE SY-MSGTY NUMBER SY-MSGNO
* WITH SY-MSGV1 SY-MSGV2 SY-MSGV3 SY-MSGV4.
ENDIF.
 
write:/ wa_output.

Regards,

Vasanth

[Former Member]

HI vasanth.. In your program You have taken <b>wa_input as type i</b>.

and u are giving value '1.000'.

the value will be converted there it self as 1 only ,integer doesn't tale decimals ,before going to call the function module it will be converted as 1..

I think you should change the data type of wa_input

Regards Rk

[anversha_s]

hi,

just execute this. u will get the idea.

data: a(16) type p decimals 4 .
 
data: b(16) type p decimals 2 .
 
data: c(30) .
 
 
a = '56.4400'.
write:/ a .
 
move: a to b .
 
write:/ b .
 
write a to c decimals 2 .
 
write:/ c .
 
the results will be
 
a = 56.4400
b = 56.44
c = 56.44

Regards

Anver

[Former Member]

Hi Vasanth,

I've tried the function module directly in se37, am not getting output. only exception ,what we have to pass in the field decimal?

Hi Aversha,

My value contains like this '5134.4500 ML' also 3215.642000 GM' .no.of zeroes is not constant so i need like this

'5143.45 ML' and '3215.624 GM'.I hope ur logic wont work with this.

any other idea?

Thanks,

Ponraj.s.

[Former Member]

data : text1(6),

text2(6),

text(13) type c value '325.2300'.

split text at '.' into text1 text2.

shift text2 right deleting trailing '0'.

shift text2 left deleting leading space.<may be spaces>

clear text.

concatanate text1 text2 into text separated by '.'.

regards

shiba dutta

[Former Member]

Execute teh code

data : val(20) type c value '6009.12300 ML',
       v1(15) type c,
       v2(6) type c.


data : in(10) type c,
       dec(6) type c,   "increase this for ur need 
       final(20) type c.
*write:/ val .

split val at ' ' into v1 v2.

*write:/ v1 .
*write:/ v2.
*
in = trunc( v1 ).

*write:/ in.
dec = FRAC( V1 ).  "this will eliminate zeros if any

*write:/ dec.


concatenate in dec+1(5) into v1.
concatenate v1 v2 into final separated by space.

write:/ final.

"ur final o/p

see if this is working for you .

regards,

vijay

[Former Member]

Raj,

Is Ur query resolved ?

Regards,

Vijay

[Former Member]

yes, my query is resolved...

sorry for poiting late.please regret...

[Former Member]

Hi Ponraj Sundarraj,

You can go through the following code.

DATA: lv_string(20) VALUE '6009.12300 ML'.

DATA: lv_length TYPE i.

DATA: lv_len TYPE i.

DATA: lv_char.

lv_length = strlen( lv_string ).

lv_length = lv_length - 4.

WHILE lv_length > 0.

lv_char = lv_string+lv_length(1).

IF lv_char EQ '0'.

lv_len = lv_length + 1.

lv_stringlv_length = lv_stringlv_len.

ELSE.

EXIT.

ENDIF.

lv_length = lv_length - 1.

ENDWHILE.

WRITE: lv_string.

Regards,

Balakrishna.N

[Former Member]

Hi,

I tested the following code segment, it works well to suit your requirement.

Just look at the code.

DATA C(20) VALUE '6009.12300 ML'.

DATA: LEN TYPE I,

FLAG TYPE I.

LEN = STRLEN( C ).

DO LEN TIMES.

LEN = LEN - 1 .

IF ( C+LEN(1) EQ '0' ) .

C+LEN(1) = ' ' .

FLAG = 1 .

ELSE.

IF FLAG = 1 .

EXIT .

ENDIF .

ENDIF .

ENDDO.

CONDENSE C NO-GAPS.

WRITE C.

Reward if helpful.

Regards,

Sandhya

[Former Member]

Answered