Try

X=INT(X*20+.5)/20



=========================
Simon Verona
Director
Dealer Management Services Ltd

T: 0845 686 2300

Sent from my iPhone

On 11 Nov 2010, at 12:42, HARIPRASATH ILANGOVAN <[email protected]> 
wrote:

> Hi,
>  
>             Please help to solve the roundoff problem. Let us take an example 
> VAR.1 = '52.6766', now the second character to the decimal value is 7 which 
> is greater than 5 so the result should be get like 52.65 and if the second 
> decimal point is 4 instead of 7 then the output should be like 52.60. So the 
> second decimal character should be roundoff between 0 and 5 only. I tried it 
> by using the substring function and i am unable to replace it to the second 
> character. Please help to proceed, coding part is given below for your 
> reference.
>  
>     PROGRAM TEST.SUBSTRINGS
>  
>     $INCLUDE GLOBUS.BP I_COMMON
>     $INCLUDE GLOBUS.BP I_EQUATE
>     VAR.1 = '52.6766'
>     VAR.2 = FIELD(VAR.1,".",2)
>     VAR.3 = SUBSTRINGS(VAR.2,2,1)
>     IF VAR.3 LT '5' THEN
>         VAR.3 = '0'
>     END ELSE
>         VAR.3 = '5'
>     END
> * From  here i am strucking, to proceed further
>     
> END
>  
>  
> 
> Thanks and Regards,
> Hari Prasath. D.I
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