Thanks for your reaction, You are right about your conclusion how I think the code should work. I am indeed trying to run the script off one server (my_domain1) and then attempting to "feed" the function a directory structure off another server (my_domain2).
I suppose the function opendir() only work for a directory at the same server. When someone can tell me what I can do best. Because I just started with PHP and I read this code in an example, I realy need some help if you tell me to use ftp. Andre "Petre Agenbag" <[EMAIL PROTECTED]> schreef in bericht news:[EMAIL PROTECTED] > The function is correct, BUT, it looks like you are trying to run this > script off one server and then attempting to "feed" the function a > directory structure off another server. > > Your code will work "as-is" if you run the script off my_domain2, but it > won't work if you run it off my_domain asking for a directory structure > located on my_domain2. > > I am not sure that opendir() can work like this, and you would probably > need to look at some remote file functions like the ftp functions... > > > On Mon, 2003-09-15 at 08:38, Andre wrote: > > Hello, > > > > For my website I use some PHP code for navigation. Therefore I use a > > directory structure which contains some navigation files the visitor can > > open. The directory structure looks like this: > > > > my_domain > > | > > - /navigation > > | - file_1.html > > | - file_2.html > > | - file_3.html > > | > > -/images > > : > > : > > > > The name of the files the visitor can open (file_1....file_3) I detect from > > reading the files in the directory "navigation". The function I use is this: > > > > $handle=opendir("./navigation/"); > > $counter=0; > > > > while ($file = readdir($handle)) { > > > > $the_type = strrchr($file, "."); > > $is_html = eregi("htm",$the_type); > > if ($file != "." and $file != ".." and $is_html) { > > $newsflash[$counter] = $file; > > $counter++; > > > > } > > } > > closedir($handle); > > rsort($newsflash); > > reset($newsflash); > > > > With some other code I can echo the different files, this code works on the > > same server. Now I want to read the navigation files from a different server > > (my_domain2), then I get an error and the name of the files are not shown. > > When I try it with Apache on my local computer an error is shown on the > > first line > > $handle=opendir("http://domain2/navigation/";); > > > > Is there an other way of getting the filenames from a different server? The > > directory structure at my_domain is equal to my_domain2. Apparently the > > function opendir is not correct, can someone help me? > > > > Thanks in advance, > > > > Andr -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
