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�
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