How to get rid of this ajax form posting issue? - asp.net-mvc-3

Question

I am posting a form to controller action. I am getting an error from the controller.
This is the behavior I want:
If error - show the right error message on the same page without any page refresh.
If success - show the success message on the page again no page refresh
what I am getting is that the page gets fully refreshed with different url and shows this:
{"status":"Failed","msg":"\u0027Name\u0027 "}
The url changes to this: ../respond/update which is the action I post to
Basically I want to catch this Failed status and display the msg inside a span.
But why it is taking me to a different page?
Here is the view:
#using (Html.BeginForm("Update", "Respond", FormMethod.Post, new { id = "frmUpdate" }))
{
//have my form here
//submit button
}
Here is the js handler that posts the form:
$('#frmUpdate').submit(function () {
//validation
$.ajax({
url: this.action,
type: this.method,
data: $(this).serialize(),
beforeSend: //show loader
success: function (result) {
alert(result.status);
},
error: function (xhr, status, error) { alert('error'); }
});
return false;
});
Here is controller:
[HttpPost]
public ActionResult Update(ResponseModel model)
{
return Json(new { status="Failed", msg = "Name" });
}
EDIT:
In Firefox it takes to differnt url.
But in IE8 I get this error from jquery itself. Interesting...
Message: Object doesn't support this property or method
Line: 28
Char: 12724
Code: 0
URI: /Content/js/external/jquery-1.6.2.min.js?v=3

Answer 1

I think you should stop the default submit behavior by doing so:
$('#frmUpdate').submit(function (e) {
e.preventDefault();
<your code here>
});
Otherwise the form submit will happen causing the behaviour you want to avoid.
Hope this helps.

Answer 2

#using (Html.BeginForm("Update", "Respond", FormMethod.Post, new { id = "frmUpdate" }))
{
//have my form here
//submit button
}
If you are using submit button than it will post the page. You need to use type button instead of submit.
It will work.
#using (Html.BeginForm("Update", "Respond", FormMethod.Post, new { id = "frmUpdate" }))
{
//have my form here
//use type button
}