list a directory in a dropdown form menu

[Anon]

Hi -
is it possible to populate a dropdown form <select> field with the output of a directory listing?

I have the listings of a directory in $ls

$dir=opendir("/path/to/dir/");
while ($ls=readdir($dir)) echo $ls;

If not, how do I explode that string listing so that each listing is on a new line?

[eelco]

Try this:

$dp=opendir($src);
while($file=readdir($dp)){
if($file!="." && $file!="..")$filelist[]=$file;
}
@sort($filelist);

echo"<select name="select"><br>";
for($m=0; $m<count($filelist);$m++){
echo "<option>$filelist[$m]</option><br>";
}
echo"</select><br>";

[Anon]

hmm, I'm not sure that's putting things in an array - it continually spews out no value, and the web page never finishes loading (the drop down menu is infinitely long with no values).

I tried to see what the array looked like with:
foreach($filelist as $val);
echo "$val \n";

and all I got was the name of the last filename in the directory.

[Anon]

just to add to that,
the directory listing has been placed in the array as written in the first part of your script, since a count() of the array lists the correct number of files in the directory.

So how come your for() and my foreach() functions are not working, but in very separate ways?

[Anon]

just changed your code to read:

for($m=0; $m<count($filelist); $m++)

so the code is now:

$src="/path/to/dir";
$dp=opendir($src);
while($file=readdir($dp)){
if (($file!=".") && ($file!="..")) {
$filelist[]=$file;
}
}

echo ("<select name=\"select\"><br>");
for($m=0; $m<count($filelist); $m++){
echo "<option>$filelist[$m]</option><br>";
}
echo"</select><br>";

Thanks for the code!

still, why this doesn't work is beyond me!
foreach($filelist as $sorted);
echo "\n $sorted \n";

[eelco]

glad it works.

I never use foreach so i can't help you with this.