I am developing a "program" and I'm trying to take in to consideration those that are to lazy to upgrade from php3 or whatever to current 🙂 so....
I have a form like so:
<input type="file" name="upload1">
<input type="file" name="upload2">
receiving script is:
if($upload1 != "")
{
copy($upload1, "./cat_images/".$upload1_name);
$lg_image = $upload1_name;
}
if($upload2_name != "")
{
copy($upload2, "./cat_images/".$upload2_name);
$sm_image = $upload2_name;
}
Is there a way to do this. I have seen lots of posts about this, but no answers other that using $_FILE[whatever] and I don't want to do this as it's only available in a later version of PHP
I can echo out $upload1 and $upload2. But $upload1_name and $upload2_name both return null.
Why? and what can I do?