Simple remote image display script help

[jch02140]

Hi,

 

I am planning to display remote images with the following code but it seems to return errors.

 

<?php
$image_dir = 'http://www.domain.com/images/' ;
$dir_handle = opendir( $image_dir );
$count = 0 ;
$display = '' ;
while( $filename = readdir($dir_handle)){
if( preg_match( '/$[a-z0-9]{4}_th\.jpg$/' , $filename ) ){
    $display .= " <img src='$image_dir$filename' /> " ;
    $count++ ;
    if( $count % 10 == 0 ){
       $count=0;
       $display .= "<br />";
    }
}
}
closedir( $dir_handle );
echo $display ;
?>

 

I am thinking might be the $image_dir cannot use address format....

 

Errors is as follow

 

Warning: opendir(http://www.foo.com/images/) [function.opendir]: failed to open dir: not implemented in /home/jch02140/public_html/test.php on line 3

 

Warning: readdir(): supplied argument is not a valid Directory resource in /home/jch02140/public_html/test.php on line 6

 

Warning: closedir(): supplied argument is not a valid Directory resource in /home/jch02140/public_html/test.php on line 16

[chintansshah]

opendir is not work for http, it works with ftp only.

 

You should find any alternative solution.

1. CURL

2. Fsocket

 

Please let me know if you are not aware of about two terms.

[jch02140]

I am not sure about CURL... but I read a little on foscket.. but I am not sure how to implement it...