Re: [PHP] Oh, the problem
Sure thing. Always love to help.
-Chris
"César aracena" <[EMAIL PROTECTED]> wrote in message
001801c23136$6748bb50$adc405c8@gateway">news:001801c23136$6748bb50$adc405c8@gateway...
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'].".";
> }
> }
>
> ---
> 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
>
>
>
>
>
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
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RE: [PHP] Oh, the problem
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'].".";
> }
> }
>
> ---
> 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
>
>
>
>
>
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
--
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To unsubscribe, visit: http://www.php.net/unsub.php
[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'].".";
}
}
---
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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