jQuery.Ajax vs Ajax.beginform with unobtrusive JavaScript

First post, please be gentle :)

I am relatively new to MVC3 and am building a webapp at work.

I have several pages with CheckBoxes that turn features on/off and when they change I submit via a jQuery Ajax call and return a json success/fail to me so I can display a message.

I have some forms with a bunch of fields that I was just submitting (not with Ajax) and checking if the modelstate is valid etc.. and redisplaying the form with messages if not. I want to do this using Ajax instead.

I have a form using Ajax.BeginForm which submits properly to the controller, the model is validated and if it has errors, i return a partial view which is replaced using the UpdateTargetId.

The thing I want to do is this... If the model was valid and the save successful, I need to return a partial view nonetheless because the UpdateTargetId will replace my form no matter what. I would like to send back a 'success' flag of some kind so I can then display a message to say 'your data was saved' or something.

OnSuccess fires if the Ajax call was successful, and doesn't care if the model was valid.

I could use jQuery.Ajax to submit the form and in the controller return the return the PartialView as well as a success or fail I think?

Can anyone tell be what the best practices are when building 'Ajax' web apps?

You may use plain javascript instead of using the Ajax.Beginform helper method to handle this.

@model ProductViewModel
<div id="divForm">
@using(Html.Beginform())
{
  @Html.TextBoxFor(x=>x.Name)
  @Html.TextBoxFor(x=>x.Location)
  <input type="submit" id="btnSave" />
}
</div>
<div id="divItem" />
<script type="text/javascript">
  $(function(){
    $("#btnSave").click(function(e){
         e.preventDefault();   // prevent the default form submit

         var form=$(this).closest("form");
         $.post(form.attr("action"),form.serialize(),function(result){
            if(result.Status=="success")
            {
              //show message and load the new partial view if needed.
              $(#divForm").hide();
              $("#divItem").html(reuslt.ViewHTML);
            }
            else
            {
               alert("Error");
            }
         });
    });
  });
</script>

Assuming you have an HttpPost action method which accepts our form subimt and return JSON data back.

[HttpPost]
public ActionResult Edit(ProductViewModel model)
{
  // here you may validate the model and save if needed and return JSON back.
  //to do  : return JSON
}

Response JSON should be in the below format.

{ "Status" : "success", "ViewHTML" : "<p>This is some markup</p>" }

This answer explains how to return ViewResult in JSON response.