Why is there no output when using a PHP list function?

Question

I want to assign values of pathinfo function to variables like so:

list($dirname, $basename, $extension, $filename) = pathinfo($path_image);
echo $dirname.$basename.$extension.$filename;

However there is no output.

But if I run only:

print_r(pathinfo($path_image));

I get output like so:

Array ( [dirname] => http://blah.com/images [basename] => image123.jpg [extension] => jpg [filename] => image123)

Comment number 2 here: php.net/manual/en/function.pathinfo.php — elclanrs
– elclanrs, Commented Sep 21, 2013 at 4:10
Put error_reporting(E_ALL); ini_set('display_errors', 1); at the top of your script and let us know what errors are you seeing now. — anupam
– anupam, Commented Sep 21, 2013 at 4:11

BlitZ · Accepted Answer · 2015-07-15 14:10:50Z

5

From php.net:

this is not really a function, but a language construct

list() only works on numerical arrays and assumes the numerical indices start at 0.

To fix that, you may try to ommit result array keys by array_values(), as mentiononed in answer of @anupam:

<?php
$values = array_values(pathinfo($path_image));
list($dirname, $basename, $extension, $filename) = $values;
?>

answered Sep 21, 2013 at 4:12

BlitZ

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Comments

anupam · Accepted Answer · 2013-09-21 04:14:40Z

2

pathinfo() returns an associative array. So, your code should work as follows:

list($dirname, $basename, $extension, $filename) = array_values(pathinfo($path_image));

answered Sep 21, 2013 at 4:14

anupam

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RhymeGuy · Accepted Answer · 2013-09-21 04:14:48Z

2

Work:

list($dirname, $basename, $extension, $filename) = array_values(pathinfo($wallpaper_image));
echo $dirname.$basename.$extension.$filename;

answered Sep 21, 2013 at 4:14

RhymeGuy

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