Re: [PHP] Why won`t this work??

2001-01-21 Thread Alexander Wagner 

bard wrote:
> Plus, this is PHP, not javascript. Variables look like this, "$UN"
> not this, "UN".

$UN is a variable.
UN might be a constant, if it is defined as one. If not, its just a 
string.

Wagner

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Re: [PHP] Why won`t this work??

2001-01-21 Thread bard 

On Sun, 21 Jan 2001, Alexander Wagner wrote:

> [EMAIL PROTECTED] wrote:
> > Hi,
> >
> > Can anyone tell me before I rip my hair out, why this won`t work??
> >
> > if (!UN)
> > {
> > printf("alert('Please enter a User
> > Name.');self.history.back();");
> > }
> >
> > TIA
> > Ade
> 
> An undefined constant will return true. You have to explicitly define 
> it as 0 to make this work.
> I guess this is becauese PHP interpretes it as a string, and all 
> strings but '' are true. It has to, because it doesn't know it is a 
> constant.
> 
> Wagner
> 

Plus, this is PHP, not javascript. Variables look like this, "$UN" not this, "UN".

Brad 


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Re: [PHP] Why won`t this work??

2001-01-21 Thread bard 

Try this...



alert('Please enter a User Name.');
self.history.back();



Cheers,

Brad

On Sun, 21 Jan 2001 [EMAIL PROTECTED] wrote:

> Hi,
> 
> Can anyone tell me before I rip my hair out, why this won`t work??
> 
> if (!UN)
> {
> printf("alert('Please enter a User 
> Name.');self.history.back();");
> }
> 
> TIA
> Ade
> 
> 



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[PHP] Why won`t this work??

2001-01-21 Thread Website4S 

Hi,

Can anyone tell me before I rip my hair out, why this won`t work??

if (!UN)
{
printf("alert('Please enter a User 
Name.');self.history.back();");
}

TIA
Ade

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