On Fri, Sep 4, 2009 at 9:48 AM, Alan<alanmand...@gmail.com> wrote: > > 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?
Sorry, I left that part out. Check out jquery-json: http://code.google.com/p/jquery-json/ You might also want to return a JSON object for all outcomes, so you don't need to check for a string ("failure","reject") *or* a JSON object. $result = array( 'result' => 'success', 'photo_id' => $newPhotoID, 'src' => PATH_FROM_DOCUMENT_ROOT, 'width' => THE_WIDTH, 'height' => THE_HEIGHT ); } else { $result = array('result' => 'failure'); } } else { $result = array('result' => 'reject'); } echo json_encode($result); You'd also be able to supply a message, if you wanted: $result = array('result' => 'reject', 'msg' => 'just because'); > 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.
