[PHP] NOT an easy upload

[César Aracena]

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




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