I have this code written by another person, the code should take the params passed as a name of file, read the file and then spits the output to the browser as for downloading the file.

this is made with a password page before. anyway, here's the code: 
#!/usr/bin/perl -w

$n = $ENV{'QUERY_STRING'};
@l = split(/=/, $n);
$file = $l[1];
if ($file) {
    undef $/;
    open (F, "secure/1/$file") or die "Cannot Open File $file\n$!\n";
    binmode (F);
    $f = <F>;    
    close (F);
    print "Content-Type: application/mpg \n";
    print "Content-Disposition: filename=$file\n\n";
    print "$f";
}
else {
    print "Content-Type: text/html\n\n";
    print "Unknown file\n";
}

[download]

as you can see, it's messy, and I want to switch it to use CGI.pm, my question is; the file type isn't always application/mpg so how do I know the type of a file I've read? or is there a common header to send for all binary file types. It's working like that, but I want to make sure it doesn't crash

I won't use this code either, I'll write my own from scratch, but I just need to know the file type thing.

thanx for your help monks out there.

In reply to how to read content-type from a file? by Anonymous Monk

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