sulfericacid has asked for the wisdom of the Perl Monks concerning the following question:
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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Re: Fix this HTML-ish image sizer
by FunkyMonk (Chancellor) on Jun 18, 2007 at 18:02 UTC |