Reading a directory and displaying its contents as an URL

xxtobirichter

Hello, I am starting to play with php and i got a webfolder where i get regulary new files. I'm looking for a command that reads out a directory and then displays the url that would be required to download the file.

<?php
$ordner = "./files";
$handle = opendir($ordner);
while ($file = readdir ($handle)) {
if($file != "." && $file != "..") {
if(is_dir($ordner."/".$file)) {
echo "/".$file."<br/>";
} else {
// kompletter Pfad
$compl = $ordner."/".$file;
echo "<a href=\"".$compl."\">".$file."</a><br/>";
}
}
}
closedir($handle);
?>

Thats what I produced up to know, it shows the name of the content and then on click one can download the file. I'm now looking for an additional feature that also displays the actual url.
I have no idea what kind of command could realize the idea may someone of you can help me?

Tobi

bradgrafelman

I'm afraid I don't understand your question. Displays the actual URL? What URL?

bogu

It depends where your script is running ...

if u are in the root and reading your dir [files], then this will do:

echo "<a href=\"{$compl}\">{$file}</a><br/>";

if u are in a subdir [script] u need to reach the root by going back (../../../):

echo "<a href=\"../../../../{$compl}\">{$file}</a><br/>";

halojoy

I have this little script.

It will read all files in same directory
and create a list with link to each file.

if you do not want it to open links in New window,
then remove target="_blank"

<?php
// directory view
$files = scandir( '.' );

echo '<body bgcolor="#DDDDDD">';
echo "<b><i>DIRVIEW FILE INDEX</i></b>";
echo "<hr>";
foreach( $files AS $name )
    if(substr($name,0,1)!=".")
        echo '<a href="'.$name.'" target="_blank">'.$name.'</a><br>';
echo '</body>';
exit;