5 replies to this topic

[xirochanh]

Hi bros,

I'm totally newbie, and this is maybe most silly question ever, but please help.
I got some PHP lines like this:

$var1 = 1;

if($var==1){

   echo 'yes';

}

I got message: "Notice: Undefined variable".
Then if I add var before $var1, error thrown is: "Parse error: syntax error, unexpected T_VAR".
If I hide notice message by statement

error_reporting (E_ALL ^ E_NOTICE);

, then no message shown, but the variable $var1 is not declared and assigned expected value - nothing is displayed.

So, what is the correct and easy way to declare a variable?
I'm running WampServer 2.1a

Thanks a lot.

[WingedPanther]

$var1 is not the same as $var

[xirochanh]

with $var1, I mean any variable name, so, when I say I add var before $var1, i mean:

var $var1 = 1;

if($var==1){

   echo 'yes';

}

[WingedPanther]

Suggestion: what is the actual code and error?

[njr1489]

Here's an answer about your keyword var: What does PHP keyword 'var' do? - Stack Overflow

The var keyword is no longer needed in PHP. PHP is a loosely typed language, meaning you don't have to declare a variable's type before using it. That being said, the keyword var is different from the variable $var. Also $var1 is not the same as $var.

[John]

$var1 = 1;

if($var==1){

   echo 'yes';

}

As the warning suggests, $var is undefined. If you want to get rid of the warning, define $var.

$var1 = 1;

$var = null;

if($var==1){

   echo 'yes';

}