Simple php question: opendir("something".$dir);?

hi there,

i’ve some code with this line:

if ($handle = opendir("http://www.base-ment.com/Flash/source/".$dir)) {

this gives this error:

Warning: opendir(http://www.base-ment.com/Flash/source/band): failed to open dir: not implemented in i:\mijn documenten\inet\basement new\flash\flash\source\pics2.php on line **6
**

[size=2]what i want is this:[/size]
[size=2]if i call file.php?dir=band from my flash file i want this in the php file:[/size]
[size=2]i**[size=2]f** ([/size][size=2][color=#800000]$handle[/color][/size][size=2] = opendir([/size][size=2][color=#0000f0]http://www.base-ment.com/Flash/source/band[/color][/size][size=2])){[/size][/size]

[size=2]who knows what’s wrong and how to correct it?[/size][size=2]
[/size]
[size=2]
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