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

[Chris]

imagesx()  requires an image resource , not a file path.  so you'd need 
to use one of the imagecreatefrom* functions to create the resource.

But it looks like all you want is the width and height of an image.... 
If so use get imagesize() ( which *does* except a pathname, not a resource)

Chris
http://www.php.net/image
http://www.php.net/getimagesize
Robert Sossomon wrote:
Here's the code for the page...
<?php
// this is the path of your files folder. Change 'pathofile'
$dir = "images/";
// this takes the files in the directory and puts them in an array
$array = array();
$handle = opendir($dir);
$extensoes = array("jpg", "png", "jpeg", "gif");
while ($file = readdir($handle))
{
 for ($i = 0; $i < count($extensoes); $i++)
 {
  if (eregi("\.". $extensoes[$i] ."$", $file))
  {
   $array[] = $file;
  }
 }
}
closedir($handle);
sort($array);
reset($array);
// 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;
// this is set up as an <img> tag and will show a random image
print "<img src=\"$path\" height=\"$height\" width=\"$width\">\n";
?>
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