On Mon, 4 Nov 2002, 1LT John W. Holmes wrote:

>How about just 
>
>print_r($_FILES);
>
>Does that return anything?

It just return the same as before, i.e. "Array ( )"

---sram

>---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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>> 
>


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