Hi thanks to Perls of Wisdom and a little help from Dmitry Ovsyanko, I'm farther down the line with my new simple script for converting xlsx2csv. Now I have a new error:

"error: No filename given at mp3.pl line 33"

Here's the code again:

#!/usr/local/bin/perl use Spreadsheet::XLSX; use Spreadsheet::XLSX::Utility2007; use Tk; my $mw = new MainWindow; my $mbar = $mw -> Menu(); $mw -> configure(-menu => $mbar); my $file = $mbar -> cascade(-label=>"File", -underline=>0, -tearoff => + 0); $file -> checkbutton(-label =>"Open", -underline => 0, -command => [\&menuopenClicked, "Open"]); $file -> command(-label =>"Save", -underline => 0, -command => [\&menusavedClicked, "Save"]); $file -> separator(); $file -> command(-label =>"Exit", -underline => 1, -command => sub { exit } ); MainLoop; sub menuopenClicked { my $typesopen = [ ['Excel 2007 files', '.xls'], ['All files', '*'],]; my $mainfilepath = $mw->getOpenFile(-filetypes => $typesop +en, -defaultextension => '.xls'); my $regions = '1-A1:G:110' ; my $rotate = 1; open($mainfilepath, "> $mainfilepath"); my $filename = $mainfilepath; close $mainfilepath; my $csvtxt = Spreadsheet::XLSX::Utility2007::xls2csv($filename, $regio +ns, $rotate); sub menusavedClicked { my $typesaved = [ ['Excel 2007 files', '.xlsx'], ['All files', '*'],]; my $saved = $mw->getSaveFile(-filetypes => $typessaved, -defaultextension => '.csv'); open($saved, "> $saved") || die "Can't create <$saved> for output!\n"; print ($saved $csvtxt) if $saved; close $saved; }}
I have tried various methods to get rid of this error, like putting single or double quotes around $filename or removing the statement:
open($mainfilepath, "> $mainfilepath");
I also tried loading an xls file instead of an xlsx file, after all the function is called xls2csv for some reason. That's the way the script is now, but it doesn't make a bit of difference. Additionally I copied both xls and xlsx files to the root, C:\ in my Windows installation thinking it might be a path issue. Neither file worked and neither file has a space in the name. It seems I am defining the filename with $filename, what's wrong?

In reply to Further On Down the Line by socrtwo

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