[SOLVED] Reading Files from a directory.

[keeps21]

I have some code which opens a directory, reads the files and assigns the file url to a variable - $image_url

 

        $baseUrl = "http://localhost";
        $gid = 1;
$dirname = "C:\\wamp\\www\\rd\\images\\galleries\\{$gid}";

$dir = opendir($dirname);

while (  false != ($file = readdir($dir)))  {
	if ( ($file != ".") and ($file != "..")  and ($file != "Thumbs.db") ) {
		$image_url = "{$baseUrl}/images/galleries/{$gid}/{$file}";
	}
}

closedir($dir);

 

This code provides me with the url of each file in the directory. I only really need to get the first file (i only need 1 file) in the directory but I'm not quite sure how to go about doing it.

 

At the minute to get just the one file I'm just overwriting $image_url; each time the loop iterates - though this isn't very efficient.

 

Any suggestions would be greatly appreciated.

[keeps21]

bump, cheers.

[Mad Mick]

Does it not matter which file it is?  If not then just lose the loop. Not sure what you mean otherwise...

 

Otherwise keep the loop but load images into array then search it for the image you want.

[keeps21]

Nope it doesn't matter which image, just any image from that directory.

 

The problem is if I remove the while loop, the first "file" in the list is . or .. rather than the first image.

 

I guess I could sort this problem by checking that the file is an image, but again that would require a loop?

[Mad Mick]

Sounds like you answered it yourself - keep looping until file is not ., .. or thumbs then break out of it.

[keeps21]

Cheers for your help.

 

Fixed code is as follows:

$baseUrl = "http://localhost";
$gid = 1;
$dirname = "C:\\wamp\\www\\rd\\images\\galleries\\{$gid}";
$dir = opendir($dirname);

while (  false != ($file = readdir($dir)))  {
    if ( ($file != ".") and ($file != "..")  and ($file != "Thumbs.db") ) {
        $image_url = "{$baseUrl}/images/galleries/{$gid}/{$file}";
        break; // break from while loop
    }
}
closedir($dir);