Rounding to the nearest 5?? 1

[Palmbak]

Does anyone know how to round a number to the nearest 5.00?

Ex. If I have 123.54, I want to round it to 125.00 or if I have 127.52 I would want that rounded to 130.00.

Thanks,

I am using Crystal Reports 8.5


Jon

 

[JoeCrystal]

Dim X as Number
Dim R as Number

X = 123.59

R = Remainder(x,5.)

If R < 2.5 Then
Formula = X - R
ElseIf R >= 2.5 Then
Formula = X + (5-R)
End If

 

[Ngolem]

I don't think that does it JC

My approach would be to truncate the number to the first place left of the decimal and do a subtraction

WhilePrintingRecords
numbervar x := {table.numbertoround};
numbervar temp;
numbervar test;

temp := truncate(x,-1);
test := x - temp;

If test > 5 then
temp := Temp + 10
else if test > 0 then
Temp := Temp + 5
else
Temp; // the case where the number like 120

Temp; Jim Broadbent

 

[krishna]

Dear Jim Broadbent

Can you explain whhy you think JC's formula will not work?

Regards
Krishna Kumar

 

[Madawc]

If I had to do such a thing, I would divide by five, then round to the nearest whole number, and finally multiply by five again.

To do this in Crystal is trickier than in a language that lets you define your own data fields. I'd suppose that you'd need a chain of formula fields to manage it. But with a bit of trial and error it should be possible.

As to whether the other methods would work, I can't comment. Madawc Williams
East Anglia
Great Britain

 

[krishna]

Dear Madawc Williams

OK. That is another good method.

But theoritically JC's formula have to work since he is finding out the remainder and adding/subtracting the remainder.

Regards
Krishna Kumar

 

[Naith]

Krishna,

In the example that Palmbak gave, he illustrated scenarios whereby the nearest five always resulted in rounding the original number UP.

If you understand &quot;round a number to the nearest 5.00&quot; (combined with him examples) to mean rounding UP, then use Ngolem's example.

Otherwise, if rounding to the nearest 5 means (as I suspect it might) rounding to the nearest number divisible by 5, regardless if the number may end up being rounded DOWN, then JoeCrystal's formula is sound.

Naith

 

[krishna]

Naith

Ok. Things are clear now. I didnt give close attention to the example.

Palmbak , Now please clarify what should happen if the no is 121.23

Thanks Naith.

Regards
Krishna Kumar

 

[Palmbak]

numberVar X:= {@%OfNatRates};
numberVar R:= Remainder(X,5.);


If R < 2.5 Then
X - R
Else if R >= 2.5 Then
X + (5-R)

This was the formula that used and it works....

Thanks JoeCrystal.

Jon Palmbak

 

[Ngolem]

Krishna

In the example given...there was no discussion about rounding to the nearest &quot;5&quot; it was either to 120 or 130....not to 120 ,125 or 130 so JC's formula would not work.

I should have perhaps explained my reason better.

I think both of our methods fail with negative numbers though on a quick glance. Jim Broadbent

 

[synapsevampire]

The first post stated round to the nearest 5, which Joe did, and very efficiently.

-k

http://www.informeddatadecisions.com

kai@informeddatadecisions.com

 

[krishna]

Hi Jim

Thanks for the explanation. Things are clear now.

Regards
Krishna Kumar

 

[Ngolem]

I suppose on re-reading I read the question wrong...sorry Jim Broadbent