Uploading image question

Dolemite50

Hi, I am trying to do an upload script for the first time and am struggling. I have created a folder on the root called bandpics. This is the full path:
home/kyle/www/bandpics

I also have a variable stored called $bandName, and I want to name the pic that variable when it is inserted into the folder. So if
$bandName = Turdz, I want the pic to be called Turdz.jpeg or whatever the extension is.

I am pretty lost by the tutes I have found so far. Heres what I have: (the input file field name is "image")

if($FILES['image'][tmp_name'] != "none"){
copy ($FILES['image']['tmp_name'], "/home/kyle/www/bandpics/".$_FILES['imagefile']['$bandName']);
or die ("Could not copy");

Thanks for ANY input on this.

gufmn

This line doesn't make sense:

copy ($FILES['image']['tmp_name'], "/home/kyle/www/bandpics/".$FILES['imagefile']['$bandName']);

$FILES is an array created for you by php as soon as the file is uploaded. The first part is ok since the name of your input is "image" as you said, so $FILES['image'] should exist. But unless you have another upload called "imagefile", $_FILES['imagefile'] won't.

Should be something like:

copy ($_FILES['image']['tmp_name'], "/home/kyle/www/bandpics/".$bandName);

I've never renamed uploaded files like this so I'm not sure but you might also have to include the file extension which you can get from $FILES['image']['type']. And if you just want to use the original file name, you can use $FILES['image']['name'].

Hope this helps.

Dolemite50

Thank you very much,
If the $bandname title will be tough, what do you recommend I do for if two people upload a file called pic001.jpeg or whatever?

gufmn

Just a thought, you could append either a session id or the temp name ($_FILES['image']['tmp_name']) to the begining of the file name. Although someone else might have a better solution.

Dolemite50

I have a thought on this.
What would be the syntax to just add the variable to the beginning of the file so that if
$bandname = turdz
$picname = 001.jpg
final result = turdz001.jpg?

gufmn

$bandname .= $picname;

means the same as:

$bandname = $bandname.$picname;