Hello,

My code is

$path = "../Data/01_OPL-277/02_SEISMIC DATA/01_2D SEISMIC/01_OLD 2D/01_2D NAVIGATION DATA/";
$dir_handle = opendir($path) or die("Error");

The result I get is "Error". I have other files with the opendir() function but they work. When I tried to echo the result of $path with the code:

$path = "../Data/01_OPL-277/02_SEISMIC DATA/01_2D SEISMIC/01_OLD 2D/01_2D NAVIGATION DATA/";
$dir_handle = opendir($path);
echo $dir_handle;

I get the result "Resource id #3".
Does anyone know what that means?