Question

I want to post a form via ajax call also model will be passed into the action method, but want to get Model errors via json. How can I do this?

Answer 1

You could write a custom action filter:

public class HandleJsonErrors : ActionFilterAttribute
{
    public override void OnActionExecuting(ActionExecutingContext filterContext)
    {
        var modelState = (filterContext.Controller as Controller).ModelState;
        if (!modelState.IsValid)
        {
            // if the model is not valid prepare some JSON response
            // including the modelstate errors and cancel the execution
            // of the action.
            // TODO: This JSON could be further flattened/simplified
            var errors = modelState
                .Where(x => x.Value.Errors.Count > 0)
                .Select(x => new
                {
                    x.Key,
                    x.Value.Errors
                });
            filterContext.Result = new JsonResult
            {
                Data = new { isvalid = false, errors = errors }
            };
        }
    }
}

and then put it into action.

Model:

public class MyViewModel
{
    [StringLength(10, MinimumLength = 5)]
    public string Foo { get; set; }
}

Controller:

public class HomeController : Controller
{
    public ActionResult Index()
    {
        return View();
    }

    [HttpPost]
    [HandleJsonErrors]
    public ActionResult Index(MyViewModel model)
    {
        // if you get this far the model was valid => process it
        // and return some result
        return Json(new { isvalid = true, success = "ok" });
    }
}

and finally the request:

$.ajax({
    url: '@Url.Action("Index")',
    type: 'POST',
    contentType: 'application/json; charset=utf-8',
    data: JSON.stringify({
        foo: 'abc'
    }),
    success: function (result) {
        if (!result.isvalid) {
            alert(result.errors[0].Errors[0].ErrorMessage);
        } else {
            alert(result.success);
        }
    }
});