Hi Tom,

You are placing the wrong thing in the source attribute:
http://code.google.com/chrome/extensions/tabs.html#method-captureVisibleTab

<http://code.google.com/chrome/extensions/tabs.html#method-captureVisibleTab>The
callback function for captureVisibleTab has this defined as:
"function(string dataUrl) {...})"

 You would use the "dataUrl" to assign the 'src' property of an HTML Image
element for display.

An example being:

<script>
chrome.tabs.captureVisibleTab(null, function(dataUrl) {
  content = document.getElementById('screenshots');
  var image = document.createElement('img');
  image.setAttribute('src', dataUrl);
  content.appendChild(image);
});
</script>

<span id="screenshots"></span>

-Mohamed Mansour


On Sat, Dec 12, 2009 at 11:55 AM, Tom <tom.extensi...@googlemail.com> wrote:

> Is it supposed to be used like this?
>
> var visualTab = chrome.tabs.captureVisibleTab(20, function(capture)
> {}););
> document.getElementById('content').innerHTML += '<img src="'+visualTab
> +'" />';
>
> Thats what I'm doing at the moment and I can't get it to work. (I'm
> sure 20 is a valid tab id)
>
> Please could someone tell me what I'm doing wrong.
>
> Thanks
>
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