PHP var construction (simple problem)

[Markp.com]

Hi people, I've forgotten how to do something in PHP and its driving me mad!

I'm trying to do a simple thing, upload images and save their paths to a database.

It all works fine, I just can't remember how to create a variable from a string.

Basically the variables going in to the Database are Img1, Img2, Img3, Img4.

My code below is a loop that writes out each image as they are uploaded.

How do I set the variable Img1 with the path of the file, inside the loop?

I can't do $Img1 = $uploadfile; as that will always write out $Img1 with each pass of the loop with the current image path.

I need to do something like
$Img.$b = $uploadfile;
$b++;
each time the loop runs, but I don't think $Img.$b will work.

Here is my code

    $uploadfile = $uploaddir . basename($_FILES['userfile']['name'][$i]);
    $ext = strtolower(substr($uploadfile,strlen($uploadfile)-3,3));
    if (preg_match("/(jpg|gif|png)/",$ext))
    {
     if (move_uploaded_file($_FILES['userfile']['tmp_name'][$i], $uploadfile))
     {
// Here I want to add the variable for each image ready to go in to a database.
		echo "Img".$imagenum." ".$uploadfile."<br>";
		$imgset = "Img".$imagenum;
		echo "<img src=\"".$uploadfile."\"><br>";
	  $success++;
	  $imagenum++;
     }

Any ideas?

[MyFriendIsATaco]

You can just use an array:

$images = array();

for($i=0;$i<$totalimages;$i++)
{
    $images[$i] = $uploadfile;
} 

That's just some rough pseudo-code, but it just shows you how to use an array.

BUT, if you really wanted to work with dynamically named variables, you could do this:

for($i=0;$i<$totalimages;$i+)
{
    $image = 'Img'.$i;
    $$image = $uploadfile;
} 

Notice the $$? It means use the value of $image as the variable name. So in this instance, you'd have access to variable names such as $Img0, $Img1, etc. But really, don't do that. Just use an array.