Oh, the problem

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 $username; //... the rest of the code } // now do the queries and everything else. I think that will fix your problem. "César aracena" <icaam@icaam.com.ar> wrote in message news:000301c23129$aebbab30$adc405c8@gateway... 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 = $userid.".jpg"; copy($_FILES['devpicture']['tmp_name'], "c:/apache/htdocs/os-seek/photos\\".$username); $username; } else { echo "Possible file upload attack: filename ".$_FILES['devpicture']['name'].".<br>"; } } ------------------------------------------- and then the updating ------------------------------------------- pic_upload($id); $query = "UPDATE os_developers SET devpicture = '$username' WHERE devid = $id"; $result = mysql_query($query) or die(mysql_error()); Thanks in advance, Cesar Aracena CE / MCSE+I Neuquen, Argentina +54.299.6356688 +54.299.4466621