Go Premium for a chance to win a PS4. Enter to Win

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 671
  • Last Modified:

Read STDIN in perl/NT

I am trying to read STDIN with the following code:  It worked on unix system but now does not work on NT.  STDIN contains a .jpg file of about 10K.  How can I get STDIN into $input??
$len = 0;
 $input = '';
while ($len != $ENV{'CONTENT_LENGTH'}) {
 $buf = '';
 $len += sysread(STDIN, $buf, $ENV{'CONTENT_LENGTH'});
 $input .= $buf;
}

0
donb1
Asked:
donb1
  • 3
  • 3
1 Solution
 
sybeCommented:
you have to use binmode on NT when it's about binary data.

like in here:

open(OUTFILE, ">upload/test.gif");
binmode (OUTFILE);
syswrite (OUTFILE, $in{'upfile'}, length $in{'upfile'});
close(OUTFILE);


0
 
donb1Author Commented:
I am trying to upload a gif file over the internet to a particular folder in NT.  I use a form:
<form action="pictur.pl" method="POST" enctype="multipart/form-data">
The gif file should be in <STDIN> along with some other form data --  How do I get it out of <STDIN>?  On a unix system, I use the code in my original question.  On an NT sysem, I only get a very small part of it (and even text is limited to about 4K).

0
 
sybeCommented:
I use cgi-lib to do that:

this is the code for the perl file:

require "./cgi-lib.pl";
$cgi_lib'maxdata = 50000;         
$ret = &ReadParse;
open(OUTFILE, ">upload/test.gif");
binmode (OUTFILE);
syswrite (OUTFILE, $in{'upfile'}, length $in{'upfile'});
close(OUTFILE);

print &PrintHeader;
print &HtmlTop("File Upload Results");

print "$in{'upfile'}";

print "\n<P>File was uploaded<P>";


print &HtmlBot;

===========

See for cgi-lib.pl:
http://www.bio.cam.ac.uk/cgi-lib/
0
Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
donb1Author Commented:
OK, that really works great.  One more question,
How can I get the name of the file that the person sent?
ie, the form field is a "file" field and contains the name of the file he uploaded.  $in{'upfile'} is the file itself.  I also need the name of the file, if possible.
Thanks
0
 
sybeCommented:
ok

In the perlscript, change

$ret = &ReadParse;

to:

$ret = &ReadParse(\%cgi_data,\%cgi_cfn,\%cgi_ct,\%cgi_sfn);

The the complete path of the uploaded file can be retrieved by:

$cgi_cfn{'upfile'}

You'll have to do some string processing before you have the actual name (because it is the complete path !).

See also http://cgi-lib.stanford.edu/cgi-lib/ex/perl5/fup.cgi.txt

(that's where I got it from)


0
 
donb1Author Commented:
Thanks very much
0

Featured Post

Hire Technology Freelancers with Gigs

Work with freelancers specializing in everything from database administration to programming, who have proven themselves as experts in their field. Hire the best, collaborate easily, pay securely, and get projects done right.

  • 3
  • 3
Tackle projects and never again get stuck behind a technical roadblock.
Join Now