[jQuery] Re: An idiot's question about returning results from a form submission

[Alan]

When you say...
        Then, in your JS function, test the returned text for
"failure" or
        "reject" first, or parse the JSON object and create a new
image tag
        with the supplied src, width, & height.

... what's the exact code for parsing the JSON object within my
current structure?

For instance, I currently submit the form and post a static dialog box
with the code...

         $('#imageUploader1').ajaxForm(function() {
                    alert("You just uploaded using Form 1");
          });

My PHP code now concludes with the following:
      echo json_encode(array ("result" => "success", "photoID" =>
$newPhotoID));

I tried replacing my static alert with the following alert (just to
make sure I'm seeing the data):

         $('#imageUploader1').ajaxForm(function() {
                    var data = eval('(' + data + ')');
                    alert("You just uploaded using Form 1 and the
result was " + data.result);
          });

Unfortunately, that's not working. What am I missing? Since I'm using
jquery and jquery form as is, I can't see how to specify the JSON info
(as I would be able to do if I was manually doing the form
submission.

Thanks, as always.