I'm positive that it will always be set, 'cause I set it. It's just
going the comparison.

if($var)

used to work for 

if($var != 0) or if($var != "0")

but that doesn't seem to work since I upgrade. So I'm just going to do

if((int)$var)

from now on, 'cause that should always work. I'll typecast everything
that being compared to and int to make sure.



On Tue, 13 Jul 2004 14:01:47 -0700 (GMT-07:00),
[EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> http://marc.theaimsgroup.com/?l=php-general&m=108646344905178&w=2
> 
> 
> 
> 
> -----Original Message-----
> From: Matthew Sims <[EMAIL PROTECTED]>
> Sent: Jul 13, 2004 12:15 PM
> To: [EMAIL PROTECTED]
> Subject: Re: [PHP] php 4.3.7/5.0
> 
> > I've noticed that in the last release of php 4.3.7 (or 5.0.0), that
> > when checking the value of a variable has changed.
> >
> > ex:
> >
> > if($var){ /* do something */ }
> >
> > doesn't work anymore. I've had to change code to
> >
> > if($var > 0)
> >
> > but the problem is, what if $var was converted to $var = "0" instead
> > of $var = 0 somewhere throughout the code that I didn't know about?
> >
> > Is there a way to be sure what's going on here?
> >
> > --
> > -Josh
> 
> What about the isset function? Or any of the following:
> 
> http://us3.php.net/manual/en/ref.var.php
> 
> --Matthew Sims
> --<http://killermookie.org>
> 
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-- 
-Josh

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