Re: imagesx(): supplied argument is not a valid Image resource (CODE)

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Robert Sossomon wrote:
<snip>
> // this is the path of your files folder. Change 'pathofile'
> $dir = "images/";

I would make this the path from root, ie...

$dir = "/root/path/images/";

<snip>
// takes the objects in the array and randomizes them
$arrayobj = array_rand($array);
$random = $array[$arrayobj];
// this creates the path name to the random file
// $path becomes the path name (duh) for the file
$path = $dir . $random;

$width = imagesx($random)/4;
$height = imagesy($random)/4;

You need to supply the path to the image...I'm betting you wanted to do...

$width = imagesx($path)/4;
$height = imagesy($path)/4;

--
John C. Nichel
ÜberGeek
KegWorks.com
716.856.9675
john@xxxxxxxxxxxx

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