Re: [PHP] Math rounding problem

[hugh danaher]

ceil() maybe
- Original Message -
From: "Charlie Killian" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Friday, February 08, 2002 4:14 PM
Subject: [PHP] Math rounding problem


> For an arbitrary large number I need to round() it up to the hundreds
place
> if it is not divisible by 100 and leave it untouched if it is.
>
> So 1100 would round to 1100 and
> 1101 would round to 1200.
>
> Is there a clean way to do this?
>
> Currently I'm:
> $scale = round($scale+49, -2); // round up to hundreds.
>
> But this is messy.
>
> Charlie
>
>
>
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RE: [PHP] Math rounding problem

[Charlie Killian]

This equation  from Bogdan is simple and working:

$scale=ceil($scale/100)*100;

Thanks to all those who helped,

Charlie


> -Original Message-
> From: Bogdan Stancescu [mailto:[EMAIL PROTECTED]]
> 
> $scale=ceil($scale/100)*100, maybe?
> 
> Bogdan

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Re: [PHP] Math rounding problem

[DL Neil]

Charlie,

> For an arbitrary large number I need to round() it up to the hundreds place
> if it is not divisible by 100 and leave it untouched if it is.
> 
> So 1100 would round to 1100 and
> 1101 would round to 1200.
> 
> Is there a clean way to do this?
> 
> Currently I'm:
> $scale = round($scale+49, -2); // round up to hundreds.
> 
> But this is messy.


You thought that was messy? Try:

if ( ( intval( $scale  / 100 ) * 100 ) < $scale )
{   
   $scale = 1 + ( intval( $scale  / 100 ) * 100 );
}

Regards,
=dn



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