Chris. Thanks for living me a hand in such a fast way. It helped me
fine.

César.

> -----Original Message-----
> From: Chris Earle [mailto:[EMAIL PROTECTED]]
> Sent: Monday, July 22, 2002 12:51 AM
> To: [EMAIL PROTECTED]
> Subject: [PHP] 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 $username;
> //... the rest of the code
> }
> // now do the queries and everything else.
> 
> I think that will fix your problem.
> 
> "César aracena" <[EMAIL PROTECTED]> wrote in message
> 000301c23129$aebbab30$adc405c8@gateway">news:000301c23129$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
> 
> 
> 
> 
> 
> 
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