My understanding is that upload() is only for input type="file" only and
that param() handles all of the other form elements. That's how I've
implemented it here and all works as expected.
BTW, I don't do any checking to see how the form was submitted. I use
Apache::Request for all of my form submission methods and everything works
fine, no matter how the information was submitted.
--
From: John S. Evans[SMTP:[EMAIL PROTECTED]]
Sent: Thursday, November 18, 1999 5:54 PM
To: modperl
Cc: [EMAIL PROTECTED]
Subject: multipart POST problems
I've been attempting to write a perl module that handles POSTs of type
multipart/form-data, and have been having a rough time.
I'm using Apache::Request to process the request. I have dumped the
content-type of the incoming request, and verified that it's
"multipart/form-data". I can use param() to get the parameters, but I
can't
seem to use upload() to access the blocks directly.
I've included the code from my test handler below. Basically it attempts
to
access a large parameter MIME block two ways - using param() and using
upload(). The param() version works fine, but the upload() version can't
find the block.
Any clues? The only thing that I can think of is that for this test case,
the MIME type of the "message" block is text/plain (it's in a TEXTAREA
field, for testing purposes). Is it possible that Apache::Request will
not
allow me to process "normal" form fields with upload()?
-jse
sub handler
{
my $r = shift;
my $apr = Apache::Request-new($r);
my $block;
my $buffer;
$r-content_type('text/html');
$r-send_http_header();
$buffer = $apr-param('message');
$r-print("bMessage:/bbrpre$buffer/prebr");
$block = $apr-upload('message');
if ($block)
{
$r-print("Found message");
}
else
{
$r-print("No message");
}
return OK;
}
