计算正确减少比例的照片尺寸在php中的宽度和高度？

Question

I have 2 variables available about a photo i'm importing, width and height.(I could use imagesx etc, but i'm importing in bulk and it takes way too long)

eg..

$w = 720;
$h = 540;

How can I workout the correct height for the photo if I say the max width is 180px, I know the correct height should be 135px (on a standard landscape photo). But the starting width and height i'm given will change depending on the photo.

Keeping the image in proportion what calculation do I need to do to workout the reduced width and height?

Final result will be in pixels so can not have any decimal places.

(I do not actually need to re-size the photo, just calculate the size.)

Thanks in advance :)

Answer 1

And this is the javascript version of @fire's answer

scaleDimensions : function(orig_width, orig_height, max_width, max_height) {
    if (orig_width < max_width && orig_height < max_height) {
        return [orig_width, orig_height];
    }

    var ratiow = max_width / orig_width;
    var ratioh = max_height / orig_height;
    var ratio = Math.min(ratiow, ratioh);
    var width = (ratio * orig_width).toFixed(0);
    var height = (ratio * orig_height).toFixed(0);
    return [width, height];
}

Answer 2

Here's one I use:

function scaleDimensions($orig_width, $orig_height, $max_width, $max_height) {
    if ($orig_width < $max_width && $orig_height < $max_height) {
        return array($orig_width, $orig_height);
    }

    $ratiow = $max_width / $orig_width;
    $ratioh = $max_height / $orig_height;
    $ratio = min($ratiow, $ratioh);
    $width = intval($ratio * $orig_width);
    $height = intval($ratio * $orig_height);
    return array($width, $height);
}

Usage:

list($width, $height) = scaleDimensions($w, $h, 180, 135);

Answer 3

720x = 540 * 180

It is a simple cross-multiplying rule. Then round the numbers to get rid of the decimal portion.