Charlie 

You're right - scrap my answer ! Was reading on my iPhone and misread the OP's 
question.

Regards

Simon 

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

T: 0845 686 2300

Sent from my iPhone

On 11 Nov 2010, at 19:06, Charlie Noah <[email protected]> wrote:

> Firstly, I think the OP's premise is incorrect. If VAR.1 = 52.6766 and he 
> wants to round to 2 decimal places, the result would be 52.68, NOT 52.65. If 
> VAR.1 = 52.6466, the result would be 52.65, not 52.60.
> 
> Second, X=INT(X*20+.5)/20 returns 52.7 (1 decimal place), and is a little 
> hard to interpret. Any time I do this, I always have to think about what to 
> multiply and divide by. I use this form:
> 
> Round to 1, then 2 decimal places (work with the OCONV version). If I goofed 
> something up here, let me know.
> 
> Code:
> VAR.1           = 52.6466
> VAR.1.ROUNDED.1 = VAR.1 1
> VAR.1.ROUNDED.2 = VAR.1 2
> X               = INT(VAR.1 * 20 + .5) / 20
> *
> CRT 'VAR.1           = ': VAR.1
> CRT 'VAR.1.ROUNDED.1 = ': VAR.1.ROUNDED.1
> CRT 'VAR.1.ROUNDED.2 = ': VAR.1.ROUNDED.2
> CRT 'X               = ': X
> *
> VAR.1           = 52.6766
> VAR.1.ROUNDED.1 = VAR.1 1
> VAR.1.ROUNDED.2 = VAR.1 2
> X               = INT(VAR.1 * 20 + .5) / 20
> *
> CRT 'VAR.1           = ': VAR.1
> CRT 'VAR.1.ROUNDED.1 = ': VAR.1.ROUNDED.1
> CRT 'VAR.1.ROUNDED.2 = ': VAR.1.ROUNDED.2
> CRT 'X               = ': X
> 
> Result:
> VAR.1           = 52.6466
> VAR.1.ROUNDED.1 = 52.6
> VAR.1.ROUNDED.2 = 52.65
> X               = 52.65
> VAR.1           = 52.6766
> VAR.1.ROUNDED.1 = 52.7
> VAR.1.ROUNDED.2 = 52.68
> X               = 52.7
> 
> Using the scaling format works every time, and no thinking. Just my $.02 
> worth.
> Charlie Noah
> Charles W. Noah Associates
> [email protected]
> 
> The views and opinions expressed herein are my own (Charlie Noah) and do not 
> necessarily reflect the views, positions or policies of any of my former, 
> current or future employers, employees, clients, friends, enemies or anyone 
> else who might take exception to them.
> 
> 
> On 11-11-2010 12:14 PM, Simon Verona wrote:
>> 
>> 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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