How about just
print_r($_FILES);
Does that return anything?
---John Holmes...
----- Original Message -----
From: "Salvador Ramirez" <[EMAIL PROTECTED]>
To: "PHP General" <[EMAIL PROTECTED]>
Sent: Monday, November 04, 2002 3:50 PM
Subject: [PHP] $_FILES
> Hi,
>
> I hope somebody could help me with a problem I have with the $_FILES
> variable.
>
> I'm trying to make a PHP script which could have access to an uploaded
> remote file. The piece of HTML is:
>
> ----------------
> <html>
> <body>
>
> <form action="ll.php" method="post" enctype="multipart/form-data">
> <input name="userfile" size=48 type="file">
> <input type="submit" name="attach" value="Add"><br>
> </form>
>
> </body>
> </html>
> ----------------
>
> and the ll.php script is:
>
> ----------------
> <?php
> printf("file=(%s)<br>", $_FILES['userfile']['name']);
> print_r(array_values($_FILES));
> phpinfo();
> ?>
> ----------------
>
> but the variable $_FILES is not set at all, because the printf doesn't
> print the name of the uploaded file nor the second line, the print_r()
> print anything for that array variable.
> So the question is: what could be happening that the variable $_FILES is
> not set?
>
> The phpinfo() function returns PHP 4.2.3, and I compiled it with the
> following configure line:
>
> './configure' '--with-mysql'
> '--with-apxs=/server/www/apache-1.3.26/bin/apxs' '--with-imap'
> '--with-imap-ssl' '--enable-track-vars' '--enable-dbase'
> '--with-pgsql=/server/postgresql'
>
> over an apache 1.3.26.
>
> Any help will be very appreciated.
>
> Thanks in advance.
>
> ---sram
> "Don't listen to what I say; listen to what I mean!" --Feynman
> Salvador Ramirez Flandes PROFC, Universidad de Concepcion, CHILE
> http://www.profc.udec.cl/~sram mailto:sram@;profc.udec.cl
>
>
>
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