Re: [PHP] $VAR = 0 != false From: Andre Fadel (fadel <email protected>)
Date: 08/07/00

Description

int empty(mixed var);

Returns false if var is set and has a non-empty or non-zero value; true
otherwise.

That means that if it's zero i will return true. You're using not (!) so
it returns false.
Stick with isset(), which is suitable for what you need. 

--
André Fadel
fadel <email protected>
Hoje E Sempre GUARANI!

On Mon, 7 Aug 2000, Brian T. Allen wrote:

> Hi *,
> 
> I have an ecommerce site where the shopping cart supports options for items.  
> 
> One of the items has an option that is stored in the database as 0 (zero).  It doesn't mean false, it means that an option was selected, and the option was 0.
> 
> If I echo out the variable, you get "0" echoed out.
> 
> $VAR = "0";
> echo $VAR;
> 
> You get what you would expect.
> 
> But, if you verify the variable exists first:
> 
> if($VAR){
>     echo $VAR;
> }
> 
> You get nothing, even though the variable is set.  It is set to zero, granted, but it is set! 
> 
> I can understand that in some types of tests you want 0 to be synonimous with false, but here I am only testing if the variable is set to a value, be it false, 0, or something else.
> 
> I just tested, and isset() works like I think it should (I am NOT the source of all truth and knowledge, just trying to make this make sense), but !empty() does not:
> 
> if(!empty($VAR)){
>     echo $VAR;
> }
> 
> even though $VAR isn't empty.
> 
> Thoughts, comments, flames?
> 
> Thanks,
> Brian
> 
> 

-- 
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: php-general-unsubscribe <email protected>
For additional commands, e-mail: php-general-help <email protected>
To contact the list administrators, e-mail: php-list-admin <email protected>