PHP var construction (simple problem)

[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.