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

From: Date: Tue, 30 Nov 2004 19:17:35 +0000
Subject: Re: imagesx(): supplied argument is not a valid Image resource (CODE)
References: 1 2  Groups: php.general 
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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@kegworks.com

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