!$subject or $subject = ""

Hi Gurus

Can I be confident that, when I write:

!$variable

This also equates to $variable == 0 or $variable = "" or should I check for those conditions separately.

I know this is a real beginner's question, but I am paranoid that I may get "false positives" or whatever you call that. Right now,
I write if(!$s || $s = "") for strings and if(!$s || $s = 0) for numbers.

Thank you in advance



- Nicolaas

Hey,

Well if you are looking to see if they are null (or not set) use
if(!isset($var))

i usually use if(!isset($var) && $var == "")

Phil



WindAndWaves wrote:
> Hi Gurus
>
> Can I be confident that, when I write:
>
> !$variable
>
> This also equates to $variable == 0 or $variable = "" or should I
check for those conditions separately.
>
> I know this is a real beginner's question, but I am paranoid that I
may get "false positives" or whatever you call that. Right now,
> I write if(!$s || $s = "") for strings and if(!$s || $s = 0) for
numbers.
>
> Thank you in advance
>
>
>
> - Nicolaas

"WindAndWaves" <access@ngaru.com> wrote in
news:0DyNd.15471$mo2.1227248@news.xtra.co.nz:

> Hi Gurus
>
> Can I be confident that, when I write:
>
> !$variable
>
> This also equates to $variable == 0 or $variable = "" or should I
> check for those conditions separately.
>
> I know this is a real beginner's question, but I am paranoid that I
> may get "false positives" or whatever you call that. Right now, I
> write if(!$s || $s = "") for strings and if(!$s || $s = 0) for
> numbers.

Yes, currently all of those conditions will evaluate the same.

Because PHP does not have strict variable typing, a variable will
evaluate false if it is unset, set to the null string, set to 0, or set
to false. There is nothing wrong with your double-checks, but you could
shorten your code to simply "if(!$s)" regardless of whether you're
checking for a set variable, a string, a number, or a boolean.

You can test using the following code:

<?php
if(!$var){
echo "No1!\n";
}
$var = '';
if(!$var){
echo "No2!\n";
}
$var = 0;
if(!$var){
echo "No3!\n";
}
$var = false;
if(!$var){
echo "No4!\n";
}

?>

Output:

$ php test.php
No1!
No2!
No3!
No4!

hth

--
Bulworth : PHP/MySQL/Unix | Email : str_rot13('f@fung.arg');
--------------------------|---------------------------------
<http://www.phplabs.com/> | PHP scripts, webmaster resources

"Senator Jay Billington Bulworth" <f@fung.arg> wrote in message news:Xns95F72C3F1D5F0CANDLETRUCK@65.24.7.50...
> "WindAndWaves" <access@ngaru.com> wrote in
> news:0DyNd.15471$mo2.1227248@news.xtra.co.nz:
>
> > Hi Gurus
> >
> > Can I be confident that, when I write:
> >
> > !$variable
> >
> > This also equates to $variable == 0 or $variable = "" or should I
> > check for those conditions separately.
> >
> > I know this is a real beginner's question, but I am paranoid that I
> > may get "false positives" or whatever you call that. Right now, I
> > write if(!$s || $s = "") for strings and if(!$s || $s = 0) for
> > numbers.
>
> Yes, currently all of those conditions will evaluate the same.
>
> Because PHP does not have strict variable typing, a variable will
> evaluate false if it is unset, set to the null string, set to 0, or set
> to false. There is nothing wrong with your double-checks, but you could
> shorten your code to simply "if(!$s)" regardless of whether you're
> checking for a set variable, a string, a number, or a boolean.
>
> You can test using the following code:
>
> <?php
> if(!$var){
> echo "No1!\n";
> }
> $var = '';
> if(!$var){
> echo "No2!\n";
> }
> $var = 0;
> if(!$var){
> echo "No3!\n";
> }
> $var = false;
> if(!$var){
> echo "No4!\n";
> }
>
> ?>
>
> Output:
>
> $ php test.php
> No1!
> No2!
> No3!
> No4!
>
> hth
>


Thanks a lot Senator...