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Home/ Questions/Q 6813263
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T20:32:58+00:00 2026-05-26T20:32:58+00:00

I need some help with something that is probably very basic. I’m working on

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I need some help with something that is probably very basic. I’m working on a PHP function that receives these possible input strings (these are examples, it could be any resolution):

1600x900
1440x900
1366x768
1360x768
1280x1024
1280x800
1024x1024
1024x768
640x960
320x480
320x480
etc

I’d like to process any one of these strings and return the appropriate aspect ratio string, in this format:

5:4
4:3
16:9
etc

Any thoughts on a simple way to approach this problem?

Edit: Here’s a reference chart I’ve been working with:

http://en.wikipedia.org/wiki/File:Vector_Video_Standards2.svg

Edit: Here’s the answer in JavaScript:

aspectRatio: function(a, b) {
    var total = a + b;
    for(var i = 1; i <= 40; i++) {
        var arx = i * 1.0 * a / total;
        var brx = i * 1.0 * b / total;
        if(i == 40 || (
                Math.abs(arx - Math.round(arx)) <= 0.02 &&
                Math.abs(brx - Math.round(brx)) <= 0.02)) {
            // Accept aspect ratios within a given tolerance
            return Math.round(arx)+':'+Math.round(brx);
        }
    }
},
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T20:32:58+00:00Added an answer on May 26, 2026 at 8:32 pm

    The following function may work better:

    function aspectratio($a,$b){
     # sanity check
     if($a<=0 || $b<=0){
        return array(0,0);
     }
     $total=$a+$b;
     for($i=1;$i<=40;$i++){
      $arx=$i*1.0*$a/$total;
      $brx=$i*1.0*$b/$total;
      if($i==40||(
           abs($arx-round($arx))<=0.02 &&
           abs($brx-round($brx))<=0.02)){
       # Accept aspect ratios within a given tolerance
       return array(round($arx),round($brx));
      }
     }
    }
    

    I place this code in the public domain.

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