amiteshpurwar Posted June 29, 2009 Share Posted June 29, 2009 Why this code is not working..?? i m a noob. Please Help me. $dir = "http://googleblog.blogspot.com/2006"; // Open a known directory, and proceed to read its contents if (is_dir($dir)) { if ($dh = opendir($dir)) { while (($file = readdir($dh)) !== false) { echo "filename: $file : filetype: " . filetype($dir . $file) . "\n"; } closedir($dh); } } ?> Link to comment https://forums.phpfreaks.com/topic/164081-about-directory-contents-in-web/ Share on other sites More sharing options...
dzelenika Posted June 29, 2009 Share Posted June 29, 2009 AFAIK $dir doesn't support http:// URL wrapper. Link to comment https://forums.phpfreaks.com/topic/164081-about-directory-contents-in-web/#findComment-865563 Share on other sites More sharing options...
Adam Posted June 29, 2009 Share Posted June 29, 2009 4.3.0 : path can also be any URL which supports directory listing I think it does, but I don't think $dir is a valid listing: http://googleblog.blogspot.com/2006 Link to comment https://forums.phpfreaks.com/topic/164081-about-directory-contents-in-web/#findComment-865566 Share on other sites More sharing options...
dzelenika Posted June 29, 2009 Share Posted June 29, 2009 On my server (windows) I got this warning: Warning: opendir(http://localhost/test) [function.opendir]: failed to open dir: not implemented in ... Link to comment https://forums.phpfreaks.com/topic/164081-about-directory-contents-in-web/#findComment-865573 Share on other sites More sharing options...
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