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Read STDIN in perl/NT

Posted on 1998-09-12
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Last Modified: 2008-02-01
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;
}

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Question by:donb1
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6 Comments
 
LVL 28

Expert Comment

by:sybe
ID: 1204838
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);


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Author Comment

by:donb1
ID: 1204839
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).

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Accepted Solution

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sybe earned 50 total points
ID: 1204840
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/
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Author Comment

by:donb1
ID: 1204841
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
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LVL 28

Expert Comment

by:sybe
ID: 1204842
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)


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Author Comment

by:donb1
ID: 1204843
Thanks very much
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