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

  1. #1

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

    hi there,

    i've some code with this line:
    PHP Code:
    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

    what i want is this:
    if i call file.php?dir=band from my flash file i want this in the php file:
    if ($handle = opendir(http://www.base-ment.com/Flash/source/band)){

    who knows what's wrong and how to correct it?

  2. #2
    i found out you can only opendir a dir on the same server... that's what is wrong with the code...

  3. #3
    what do you want to do with your script?

  4. #4
    thanks for you offer to help me... but i already solved the problem

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