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Asked: 2026-05-10T17:11:39+00:00

If I load the nextimg URL manually in the browser, it gives a new

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If I load the nextimg URL manually in the browser, it gives a new picture every time I reload. But this bit of code shows the same image every iteration of draw().

How can I force myimg not to be cached?

<html>   <head>     <script type='text/javascript'>       function draw(){         var canvas = document.getElementById('canv');         var ctx = canvas.getContext('2d');         var rx;         var ry;         var i;          myimg = new Image();         myimg.src = 'http://ohm:8080/cgi-bin/nextimg'          rx=Math.floor(Math.random()*100)*10         ry=Math.floor(Math.random()*100)*10         ctx.drawImage(myimg,rx,ry);         window.setTimeout('draw()',0);       }     </script>   </head>   <body onload='draw();'>     <canvas id='canv' width='1024' height='1024'></canvas>   </body> </html>
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1 Answer

  1. That actually sounds like a bug in the browser — you could file at http://bugs.webkit.org if it’s in Safari or https://bugzilla.mozilla.org/ for Firefox. Why do i say potential browser bug? Because the browser realises it should not be caching on reload, yet it does give you a cached copy of the image when you request it programmatically.

    That said are you sure you’re actually drawing anything? the Canvas.drawImage API will not wait for an image to load, and is spec’d to not draw if the image has not completely loaded when you try to use it.

    A better practice is something like:

        var myimg = new Image();     myimg.onload = function() {         var rx=Math.floor(Math.random()*100)*10         var ry=Math.floor(Math.random()*100)*10         ctx.drawImage(myimg,rx,ry);         window.setTimeout(draw,0);     }     myimg.src = 'http://ohm:8080/cgi-bin/nextimg'

    (You can also just pass draw as an argument to setTimeout rather than using a string, which will save reparsing and compiling the same string over and over again.)

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