Get an image's apect ratio

[reeksy]

Hi.

I need to discover an image's aspect ratio.

This is what i've got so far:

$img = dirname(__FILE__) . DIRECTORY_SEPARATOR . 'image1.JPG';

$size = getimagesize($img);

// width divided by its height
$width = $size[0];
$height = $size[1];

$ratio = $width / $height;
echo $ratio;

Which outputs

1.3297872340426 

However i need to view the ratio in the x:y format (e.g 4:3).

How do i do this!?

Regards,
Jon

[bretticus]

At a glance, my suggestion is to do the math as you are doing and use something like the round function:

<?php
echo round(1.3297872340426, 1);
?>

equals 1.3. Since you know that 1.3 is 4:3 (or closest to 4:3) you could surmise that the source image has 4:3 aspect ratio. Other sizes could be handled similarly. (yes, I agree a purely mathematical solution would be more elegant.)

[bradgrafelman]

I was curious myself as to how you might go about doing this, so I whipped up a function and it appears to work... not sure how accurate it is, though.

function get_aspect_ratio($x, $y) {
	$r = array( // array of common ratios
		array(4, 3),
		array(3, 2),
		array(16, 9),
		array(1.85, 1),
		array(2.39, 1)
	);

for($i = 0, $prev = null; isset($r[$i]); $i++)
	if($prev == null)
		$prev = abs(($x / $y) - ($r[$i][0] / $r[$i][1]));
	else {
		$next = abs(($x / $y) - ($r[$i][0] / $r[$i][1]));

		if($next > $prev)
			break;
		else
			$prev = $next;
	}

if($i == count($r))
	return false;
else
	return $r[$i-1];
}

It could be used as such:

$ratio = get_aspect_ratio($image_x, $image_y);

echo "The ratio of the image is $ratio[0]:$ratio[1]";