jQuery returning "parsererror" for ajax request

Question

Been getting a "parsererror" from jquery for an Ajax request, I have tried changing the POST to a GET, returning the data in a few different ways (creating classes, etc.) but I cant seem to figure out what the problem is.

My project is in MVC3 and I'm using jQuery 1.5 I have a Dropdown and on the onchange event I fire off a call to get some data based on what was selected.

Dropdown: (this loads the "Views" from the list in the Viewbag and firing the event works fine)

@{
    var viewHtmls = new Dictionary<string, object>();
    viewHtmls.Add("data-bind", "value: ViewID");
    viewHtmls.Add("onchange", "javascript:PageModel.LoadViewContentNames()");
}
@Html.DropDownList("view", (List<SelectListItem>)ViewBag.Views, viewHtmls)

Javascript:

this.LoadViewContentNames = function () {
    $.ajax({
        url: '/Admin/Ajax/GetViewContentNames',
        type: 'POST',
        dataType: 'json',
        data: { viewID: $("#view").val() },
        success: function (data) {
            alert(data);
        },
        error: function (data) {
            debugger;
            alert("Error");
        }
    });
};

The above code successfully calls the MVC method and returns:

[{"ViewContentID":1,"Name":"TopContent","Note":"Content on the top"},
 {"ViewContentID":2,"Name":"BottomContent","Note":"Content on the bottom"}]

But jquery fires the error event for the $.ajax() method saying "parsererror".

Answer 1

I recently encountered this problem and stumbled upon this question.

I resolved it with a much easier way.

Method One

You can either remove the dataType: 'json' property from the object literal...

Method Two

Or you can do what @Sagiv was saying by returning your data as Json.

---

The reason why this parsererror message occurs is that when you simply return a string or another value, it is not really Json, so the parser fails when parsing it.

So if you remove the dataType: json property, it will not try to parse it as Json.

With the other method if you make sure to return your data as Json, the parser will know how to handle it properly.