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 $username;
//... the rest of the code
}
// now do the queries and everything else.
I think that will fix your problem.
"C�sar aracena" <[EMAIL PROTECTED]> wrote in message
000301c23129$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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