Question

I want to echo something if the subdomain is not “mobile” and the file name is not “design” or “photo”, so echo if (not “mobile” and not “design”) or (not “mobile” and not “photo”).

The code I wrote is:

<?php
    if ( ($subdomain != mobile) && ($currentFileName != design || $currentFileName != photo) ) {
        echo ('code')
    }
?>

but it doesn’t like the compound statement. if i just do ($subdomain!=mobile) it works; if i just do ($currentFileName!=design) it works; and if i just do ($currentFileName!=photo) it works….
The above code doesn’t result in an error…it just doesn’t do anything.

Is there a particular way to compound/group the statement that I’m just not getting? I feel like this shouldn’t be difficult (and I’ve done much more complex things with php without this much trouble…).
I modelled my statement after this: http://www.php.net/manual/en/control-structures.if.php#101724

Thanks!

EDIT:
fyi:

$subdomain = implode(array_slice(array_reverse(explode('.', $baseURL)),2)); //resolves to "test"
$basename = basename($_SERVER['REQUEST_URI']); //resolves to "design.php"
$currentFileName = basename($basename, ".php"); //resolves to "design"

the new problem is that everything in the subdomain “mobile” is now being caught, so no pages in it are getting the “code” from echo (even the ones not ‘design’ or ‘photo’).

the most recent code I tried was:

if ( ($subdomain != 'test' && $currentFileName != 'design') && ($subdomain!='test' && $currentFileName != 'photo') ) {
    echo ('ajax code');
}

but still no dice 🙁

Answer 1

There’s a flaw in your logic. You most likely want && not ||. If you think about the statement for a minute it will become apparent. Let’s say the following scenario: $currentFileName is ‘design’. It will pass since it’s not ‘photo’. If we make $currentFileName equal ‘photo’ it will still pass since it doesn’t equal ‘design’.

EDIT:
And PLEASE use quotes when comparing strings. PHP will attempt to find a constant with that name first, then assume it’s a string because it doesn’t find one.