$args in an if()

[twlack]

This does what I want:
[php]if($form_lname == strtolower ($lastname) AND $form_fname == strtolower ($firstname)) {
echo ‘

’. $user->display_name . ‘

’;
}[/php]

But this does not:
[php]$args=’$form_lname == strtolower ($lastname) AND $form_fname == strtolower ($firstname)’;
if($args) {
echo ‘

’. $user->display_name . ‘

’;
}[/php]

Wat am I missing?

Thanks,
Tim

[g0dzuki99]

Doesn’t work like that in conditional statements… your ‘if’ is basically saying “if this is a string, do something” instead of evaluating the contents of the string, and will always evaluate to TRUE.

Make sense?

Just as a simpler example:

[php]
$x = 3;
$args = “$x > 5”;

if ($args) {
echo $x . ’ is greater’;
} else {
echo $x . ’ is less’;
}

//output: 3 is greater

[/php]