It appears that $username isn't being defined within the scope of the query.
You define it only in the function.

Try do this instead:

$username = "";

function pic_upload($userid)
global $username;
//... the rest of the code
// now do the queries and everything else.

I think that will fix your problem.

"CÚsar aracena" <[EMAIL PROTECTED]> wrote in message
Hi all. I'm trying to handle a picture upload. So far, I've made my
script store it where I want with the name I want. The only problem now,
is that it should store that given name into a MySQL DB table. I tried
it by calling a function which stores the picture and returns a variable
called $picname then use an UPDATE statement but that variable isn't
passed. any ideas? The code looks like this:

function pic_upload($userid)
if (is_uploaded_file($_FILES['devpicture']['tmp_name']))
$filename = $_FILES['devpicture']['tmp_name'];

$realname = $_FILES['devpicture']['name'];

$username = $userid.".jpg";


echo "Possible file upload attack: filename

and then the updating


$query = "UPDATE os_developers SET devpicture = '$username' WHERE devid
= $id";
$result = mysql_query($query) or die(mysql_error());

Thanks in advance,

Cesar Aracena
Neuquen, Argentina

PHP General Mailing List (
To unsubscribe, visit:

Reply via email to