You're right, I'm calling this via inside a OO Perl Module, its a simple wrapper to the DB_File module, but it also does a bunch of other things. This is for a simple web-based database admin screen. Info from a form is sent to a script, that calls this Module kinda like

 
my $input = CGI -> new(); 

my $image = $input -> param('image'); 
# some more form fields fetched through param()...

my $info = new -> JustinDb(); 
   $info       ->savethis(-name => 'image', 
                         -value => $image);
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the _file_upload() is always internal. This module is one of many, and the module itself 'knows' what kind of form field the info was sent from. Yes I'm tricky :) That part is checking out fine. it looks to see if this particular info is handed to it from a fileupload field, and if it is, saves the file to the $img_dir variable.

I've been fiddling with this some more, and filenames called 1photo.gif will work, but photo.gif will not work! this is incredibly weird. This is my problem and I'm still scratchin my head.

 

-justin simoni
!skazat!

A couple of things could be going wrong.

Bizarrely-worded description of CGI upload and filehandles deleted. See Re (tilly) 2: YAF(ile)U(ploading)Q(uestion) for details.

I'm assuming that you're calling your function like so:

use CGI;

my $query = CGI->new();
&_upload_file($query->param('my_upload_field'));
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use CGI;

my $query = CGI->new();
my $uploaded_filehandle = $query->upload('my_upload_field')
    or print_error_page("Sorry, you must upload a file...");
&_upload_file( $uploaded_filehandle );
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That way, you'll pass only a filehandle, and not have to worry about the vagaries of packages. The only problem is that it will not preserve the original name.

So, you might want to write your function like so:

use CGI;
use File::Copy;

## in main...
my $query = CGI->new();
my $uploaded_filehandle = $query->upload('my_upload_field')
    or print_error_page("Sorry, you must upload a file...");
my $filename = $query->param('my_upload_field');
&_upload_file( $uploaded_filehandle, $filename );


sub _upload_file {
    my ($uploaded_fh, $orig_filename) = @_;

    my $dest_filename = $orig_filename;
    $dest_filename =~ s{ ^.* ( \\ | \/ ) }{}x;
    copy($uploaded_fh, $dest_filename) or die "Couldn't copy file to '
+$dest_filename': $!; stopped";
}
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Normally, you would NEVER want to allow user input to determine the name of a filename in a CGI script, but since it's an internal application we'll let it go this once.

Finally, you might add

use CGI::Carp 'fatalsToBrowser';
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Update: Previously, I didn't understand the question thoroughly enough before posting... my apologies.
Additional update:tilly pointed out that my wording was so confusing as to be altogether wrong.

stephen

Actually CGI does not name the file what it is named in the filename. (At least it shouldn't be, and doesn't do it instead.) Instead what it does is returns a glob. If you use the glob as a string, you will see the name of the file that they are trying to upload. If you use it as a filehandle, you will be reading from the file.

And yes, I was confused for quite a while about how it worked. :-)

use CGI;
$cgi=new CGI;

$file=$cgi->param('file_field');

$newName=&FileUpload($file);

sub FileUpload{
        my $FileName=shift @_;
        my $tmpdir ="/your/path"
        my $bufferSize=4096;
        my $limit=10000000;#File Size Limit
        my ($debug, $FILE, $file, $size, $tmppath, $line, $buffer);
        $size = 0;
        if (length($FileName) != 0){
                $| = 1;
                $FILE = $FileName;
                ($file = $FILE) =~ s/^.*?([^\\\/]+)$/$1/;
                $tmppath = "$tmpdir\\$file";
                open (OUT, ">$tmppath") or print "Error opening $tmppa
+th: $!<br>\n";
                binmode OUT;
                while(($line = read ($FileName, $buffer, $bufferSize))
+ && ($size < $limit)){
                        print OUT $buffer;
                        $size += $line;
                        if ($size > $limit){
                                $err_flag = 1;
                        }
                }

                close OUT;
                $| = 0;
                $FileName = $file;
        }
        else{##Empty File
                $FileName = "";
        }       

        #Upload Error
        if ($err_flag == 1){
                unlink ($tmppath);
                $FileName = "";
        }
        $FileName;
}
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here's an update.

I'm going to stick this under 'Weird Perl Oddity' until I figure out what's going on. My guess is that passing the magic variable that acts like a filehandle in some circumstances, and a plain scalar in others to an outside module that then passes it to a private subroutine is way too much to ask for and to keep the magic. I still don't understand why a filename uploaded that has a number in its name works (1photo.gif), but a filename that has only letters (hello.gif) doesn't. I'm thinking it's something to do with the tmp filename, but I may be way off on this.

The way I solved this problem is to stick the fileupload subroutine in the script that gets called, instead of calling an OO object from the script that does the file upload stuff. Why this makes a difference, I have no idea *shrugs* I created a new CGI object just for the fileupload function, since I read that mixing the OO and regular styles of writing with the CGI.pm can muck up magic thingies.

This resolution totally breaks my abstaction of a pretty darn intricate system (probably the hardest I've done) and plunks in code that may need to be repeated in more than one script since its now outside the OO Module.

*grumble grumble grumble*

Thought everyone would appreciate a resolution, as many times I tried to find an answer to my questions, I find instead an unfinished thread.

I'm still looking for a way to put a file upload script in an outside module and maybe some wisdom on why it makes a difference where this code is anyways. I'm thinking that I need to somehow pass the filehandle to the outside (in the module) , private upload function, instead of a magic variable.

thanks, everyone for your help

-justin simoni
!skazat!