Warning: open dir: not implemented

Question

I\'m new to PHP, and I\'m trying to build a script. When I load the script, I get the following error:

Warning: opendir(http://www.hetweerinboskamp.nl/voorpagina/mov

Answer 1

opendir operates on directories on a filesystem, not HTTP URIs.

While some HTTP URIs return directory listings (the one you are using doesn't, it is a 404 error), those listings as HTML documents generated by the webserver and are not actual directories.

Answer 2

opendir() is used to open a local directory and since PHP 5.0.0 on an ftp directory.

If your PHP code runs on www.hetweerinboskamp.nl then /voorpagina/movies is actually a local directory and you can do this:

$dir ='<wwwroot>/voorpagina/movies';
if ($handle = opendir($dir)) {

where wwwroot is the root of the filesystem as seen by your php code.

If you're trying to download content from another website, try e.g. file_get_contents(). Note that if the remote server lists the content of a directory the listing is in fact an HTML page generated on the fly by the server. You may find yourself needing to parse that page. A better approach is to check whether the server offers some sort of API where it sends back the content in a standardized form, e.g. in JSON format.

Answer 3

Most remote servers does not send a directory listing back as such opendir cannot understand what your trying to do so it cant work.

You will need to use something like ftp, here is an example: http://php.net/manual/en/ftp.examples-basic.php or cURL

Answer 4

Manual claims this function works with URL's, however, it appears it doesn't.

Use a local path (either relative or absolute). For example, './voorpagina/movies'. This has solved a similar problem to me before. I hope this helps.