Getting all file names in a directory

[Lamez]

How can I get all the file names in a directory?

 

The reason is because I am using Java script to load images from a drop down box, but the you have to enter the image names, but I have tons and tons of images in my Helmet directory, so is there a for loop or something to find all the image names?

 

<form name="<?php echo $path; ?>/main/style/img/Helmets/"><p>
<select name="picture" size="1" onChange="showimage()">
<option value="img.gif">Picture 1</option>
<option value="img_2.gif">Picture 2</option>
<option value="img_3.gif">Picture 3</option>
</select>
</form>

[ratcateme]

have a look at http://nz2.php.net/function.opendir

 

Scott.

[Lamez]

Hi, and thank you for the link I did not know that existed (not the manual).

 

But could you walk me trough this example code? I have commented some lines

 

<?php
$dir = "/etc/php5/";

// Open a known directory, and proceed to read its contents
if (is_dir($dir)) {
    if ($dh = opendir($dir)) { //what is $dh?
        while (($file = readdir($dh)) !== false) {
            echo "filename: $file : filetype: " . filetype($dir . $file) . "\n"; //what is $file, filename? 
        }
        closedir($dh); 
    }
}
?>

[JasonLewis]

Your setting $dh to directory resource handle. It's in an if statement, because it can return false if it fails to open the directory.

 

For $file, if you look in the loop it's setting $file to the name of a file. So basically the loop is looping through all the files and after each loop sets $file to the new filename.

[Lamez]

oh cool, well now I have ran into a problem, I have some sub directorys, is there a way to go into the folder, and find all the images? then do that for all sub dirs?

 

I think I would need a nested loop. 

[JasonLewis]

I prefer glob, so here is one way.

 

<?php
$allowed = array("jpg", "jpeg", "gif", "png"); //allowed file extensions

$subDir = array();
$files = array();

foreach(glob("./dir/*") as $f){
if(filetype($f) == "dir"){
	foreach(glob($f."/*") as $s){
		if(filetype($s) == "file"){
			$ext = explode(".", basename($s));
			if(in_array($ext[count($ext)-1], $allowed)){
				$files[] = $s;
			}
		}
	}
}elseif(filetype($f) == "file"){
	$ext = explode(".", basename($f));
		if(in_array($ext[count($ext)-1], $allowed)){
			$files[] = $f;
		}
}
}

echo "<pre>";
print_r($files);
echo "</pre>";
?>

[Lamez]

wow that helps, but I get an output like:

 

[0] => ../../main/style/img/Helmets/Atlantic Coast Conference/Boston College.jpeg

 

how could I make it  just Boston Collage

[JasonLewis]

When you echo them out, do it like this:

 

echo basename($files[0]);

 

Or, you can change the code to this:

 

<?php
$allowed = array("jpg", "jpeg", "gif", "png"); //allowed file extensions

$subDir = array();
$files = array();

foreach(glob("./dir/*") as $f){
if(filetype($f) == "dir"){
	foreach(glob($f."/*") as $s){
		if(filetype($s) == "file"){
			$ext = explode(".", basename($s));
			if(in_array($ext[count($ext)-1], $allowed)){
				$files[] = basename($s);
			}
		}
	}
}elseif(filetype($f) == "file"){
	$ext = explode(".", basename($f));
		if(in_array($ext[count($ext)-1], $allowed)){
			$files[] = basename($f);
		}
}
}

echo "<pre>";
print_r($files);
echo "</pre>";
?>

 

I wouldn't recommend the last option, because it's good to have the directory in case you need to link it. I'd just use basename() on the file you want.

 

Actually, here is a better version.

<?php
$allowed = array("jpg", "jpeg", "gif", "png"); //allowed file extensions

$subDir = array();
$files = array();

foreach(glob("./dir/*") as $f){
if(filetype($f) == "dir"){
	foreach(glob($f."/*") as $s){
		if(filetype($s) == "file"){
			$ext = explode(".", basename($s));
			if(in_array($ext[count($ext)-1], $allowed)){
				$files[] = array($s, basename($s, ".".$ext[count($ext)-1]));
			}
		}
	}
}elseif(filetype($f) == "file"){
	$ext = explode(".", basename($f));
		if(in_array($ext[count($ext)-1], $allowed)){
			$files[] = array($f, basename($f, ".".$ext[count($ext)-1]));
		}
}
}

echo "<pre>";
print_r($files);
echo "</pre>";
?>

[Lamez]

I am a bit confused by your code.

 

could I do something like this:

 

<?php
function getExt($str) {

       $i = strrpos($str,".");
       if (!$i) { return ""; }

       $l = strlen($str) - $i;
       $ext = substr($str,$i+1,$l);

       return $ext;

}
$dir = $path."main/style/img/Helmets";

if (is_dir($dir)) {
   if ($dh = opendir($dir)) {
       while (($file = readdir($dh)) !== false) {
        if($file !== (".")){
           if($file !== ("..")){
             if($file !== ("index.html")){
                echo'<option value="'.$file.'">'.$n_file.'</option>';
                 }//close index
             }//close .
          }//close .
       }//close while loop
       closedir($dh);
   }//close open dir $dh
}//close is dir
?>

 

but then all I get listed is dir, could I add this

 

<?php
if(is_dir($file){
echo "is dir?";
}else{
echo "is file?";
?>

[Guest Xanza]

Lamez... Google is your friend; so is php.net. Don't simply keep firing off questions, it get's annoying. If you're going to ask a question make sure you have the willingness to lookup side questions that you have. Otherwise it just get's tedious for anyone to help anyone - cause they have to keep re-explaining the basics over and over.

 

[0] => ../../main/style/img/Helmets/Atlantic Coast Conference/Boston College.jpeg

 

That's 90% of what you're trying to do. So now would be a great opportunity for you to use google and lookup how to strip filenames, or even to make dynamic PHP links.

[Lamez]

that is not the problem, I am here to learn, and I can strip the file names, but I want to make the code my own, so I know exactly how it works.

 

I just figured it would be easier to do a nested loop. And I am asking if my code would be correct.

[Guest Xanza]

I just figured it would be easier to do a nested loop. And I am asking if my code would be correct.

 

Ahh, sorry. Misunderstanding on my part.

 

Nested loop... No, but the overall code would indeed be correct in the sense that it would work. There really is no such thing as "correct" in programming. If you can get it to work, it works; nothing else really matters.

[GingerRobot]

If you can get it to work, it works; nothing else really matters.

 

Err...Efficiency not in your dictionary then? Not to mention how easy something is to maintain etc...

 

To answer your question Lamez, a nested loop will only work if you have a known level of subdirectories. For example, projectfear's code (which does indeed use a nested loop) is only able to retrieve files from directories one level deep. For this to work with a more complex structure, you'll need a recursive function.

[Lamez]

well, so what would be the best way? It looks like I am going to use the code posted before.

[JasonLewis]

Yeah my code would only take you to sub-directories, not sub-sub-directories (is that what you call them? ).

I'm not sure about going any deeper, without knowing what directories there are. I'm sure I could if I sat down and tried to program it.

[JasonLewis]

Okay, not very good with recursive functions but you can try this.. It works for unlimited amounts of sub-directories, I hope.

 

<?php

$subDir = array();
$files = array();

function getFiles($dir){
$allowed = array("jpg", "jpeg", "gif", "png"); //allowed file extensions

if(substr($dir, -1, 1) != "/"){
	$dir .= "/";
}
foreach(glob($dir."*") as $f){
	if(filetype($f) == "dir"){
		$files[] = getFiles($f);
	}elseif(filetype($f) == "file"){
		$ext = explode(".", basename($f));
			if(in_array($ext[count($ext)-1], $allowed)){
				$files[] = array($f, basename($f, ".".$ext[count($ext)-1]));
			}
	}
}

return $files;
}

$files = getFiles("./dir/");
echo "<pre>";
print_r($files);
echo "</pre>";

[Guest Xanza]

Err...Efficiency not in your dictionary then? Tongue Not to mention how easy something is to maintain etc...

 

Effi... Wha?