CODEWITHSUNDEEP
jqueryasp.net-mvc-3knockout.js
Jake Pearson
Jake Pearson

Reputation: 27717

Submit json to MVC3 action

I have a form created with Knockout.js. When the user presses the submit button, I convert the viewmodel back in a model and am trying to submit to the server. I tried:

ko.utils.postJson(location.href, ko.toJSON(viewModel));

But the object was blank when it hit the server. I switched to this code:

$.ajax({
    url: location.href, 
    type: "POST",
    data: ko.toJSON(viewModel),
    datatype: "json",
    contentType: "application/json charset=utf-8",
    success: function (data) { alert("success"); }, 
    error: function (data) { alert("error"); }
});

That gets the data to the server with the correct data in it.

But what I would like is to have the data submitted so my controller can redirect to the correct view. Any suggestions?

Upvotes: 5

Views: 5279

Answers (2)

user819852
user819852

Reputation: 1

Also, it's in the example mentioned, but you may need to alter your JavaScript call to something like:

ko.utils.postJson(location.href, { viewModel: this.viewModel });

Upvotes: -2

RP Niemeyer
RP Niemeyer

Reputation: 114792

Steve Sanderson has an older sample that demonstrates getting submitted JSON data to be bound properly in your controller action here: http://blog.stevensanderson.com/2010/07/12/editing-a-variable-length-list-knockout-style/

The gist of it is that he creates an attribute called "FromJson" that looks like:

public class FromJsonAttribute : CustomModelBinderAttribute
{
    private readonly static JavaScriptSerializer serializer = new JavaScriptSerializer();

    public override IModelBinder GetBinder()
    {
        return new JsonModelBinder();
    }

    private class JsonModelBinder : IModelBinder
    {
        public object BindModel(ControllerContext controllerContext, ModelBindingContext bindingContext)
        {
            var stringified = controllerContext.HttpContext.Request[bindingContext.ModelName];
            if (string.IsNullOrEmpty(stringified))
                return null;
            return serializer.Deserialize(stringified, bindingContext.ModelType);
        }
    }
}

Then, the action looks like:

    [HttpPost]
    public ActionResult Index([FromJson] IEnumerable<GiftModel> gifts)

Now, you could use ko.utils.postJson to submit your data and respond with an appropriate view.

Upvotes: 11

Related Questions