asked on

PHP Basename Question

The code is pretty simple, but yet it still prints false... here is the live link..

http://www.tornadoeskick.com/tonylaubach/codeTest.php

Obviously the filename is codeTest.php, but it won't display true for some reason? Any suggestions?

<?php
$myURL=basename(basename($_SERVER['REQUEST_URI']));

echo "$myURL";
?>
<br><br>
<?php
if ( $myURL == codeTest.php ) {
	echo "The if statement evaluated to true";
} else {
	echo "The if statement evaluated to false";
}

?>

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Lee Wadwell

try adding to the top of your script
error_reporting(E_ALL);
and quotes around the script name.

ASKER CERTIFIED SOLUTION

g3nu1n3

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Dave Baldwin

@g3nu1n3 is right. Without the quotes, PHP looks for a constant by that name and of course it doesn't find it.

Ray Paseur

Grab a copy of this book. It will not make you a pro, but it will give you enough of a foundation in the basics of PHP to avoid simple syntax errors like this one.
http://www.sitepoint.com/books/phpmysql5/

Sidebar note: You only need to call the basename() function once.

Second sidebar note: Learn about this function:
http://us2.php.net/manual/en/function.var-dump.php

Best of luck with your PHP adventures, ~Ray

laubacht

ASKER

The quotes did it... I appreciate the fast help on that! Working a website without using Wordpress or something along those lines, so doing a lot of this from scratch! Thanks guys!