http://qs1969.pair.com?node_id=621836

sulfericacid has asked for the wisdom of the Perl Monks concerning the following question:

This was posted as an untested solution to a SOPW a week or so past and it was stated by a monk that there was indeed an error in the calculations. Can someone provide a working solution keeping my methods thus far intact? I know it can be done via Image::Image or another module but those aren't available in my workspace and I'd like to see how close I got to the final solution.

The question was: the user needed a way to determine the h/w of an image and output an html-resized version of it so that both h/w are less than a max value.. AND the image has to stay the same aspect ratio.

#!/usr/bin/perl use warnings; use strict; my $height = "1000"; my $width = "1234"; my $maxh = "600"; my $maxw = "800"; if ($height > $maxh || $width > $maxw) { my $hratio = $height / $maxh; my $wratio = $width / $maxw; my $newh = $height / $hratio; my $neww = $width / $hratio; print "Height is now $newh, Width is now $neww"; } RESPONSE FROM ALMUT: It's not quite as trivial as it might seem at first (most of us who ha +ve tried probably got it wrong the first time :) Just as an example, take a width of 1000 and a height of 100. Your al +gorithm would compute 6000x600 (W x H), although the expected result +would be 800x80 (i.e. max width doesn't exceed 800 and aspect ratio s +tays at 10:1)


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sulfericacid