How to post an array of complex objects with JSON, jQuery to ASP.NET MVC Controller?

I've found an solution. I use an solution of Steve Gentile, jQuery and ASP.NET MVC – sending JSON to an Action – Revisited.

My ASP.NET MVC view code looks like:

function getplaceholders() {
        var placeholders = $('.ui-sortable');
        var results = new Array();
        placeholders.each(function() {
            var ph = $(this).attr('id');
            var sections = $(this).find('.sort');
            var section;

            sections.each(function(i, item) {
                var sid = $(item).attr('id');
                var o = { 'SectionId': sid, 'Placeholder': ph, 'Position': i };
                results.push(o);
            });
        });
        var postData = { widgets: results };
        var widgets = results;
        $.ajax({
            url: '/portal/Designer.mvc/SaveOrUpdate',
            type: 'POST',
            dataType: 'json',
            data: $.toJSON(widgets),
            contentType: 'application/json; charset=utf-8',
            success: function(result) {
                alert(result.Result);
            }
        });
    };

and my controller action is decorated with an custom attribute

[JsonFilter(Param = "widgets", JsonDataType = typeof(List<PageDesignWidget>))]
public JsonResult SaveOrUpdate(List<PageDesignWidget> widgets

Code for the custom attribute can be found here (the link is broken now).

Because the link is broken this is the code for the JsonFilterAttribute

public class JsonFilter : ActionFilterAttribute
{
    public string Param { get; set; }
    public Type JsonDataType { get; set; }
    public override void OnActionExecuting(ActionExecutingContext filterContext)
    {
        if (filterContext.HttpContext.Request.ContentType.Contains("application/json"))
        {
            string inputContent;
            using (var sr = new StreamReader(filterContext.HttpContext.Request.InputStream))
            {
                inputContent = sr.ReadToEnd();
            }
            var result = JsonConvert.DeserializeObject(inputContent, JsonDataType);
            filterContext.ActionParameters[Param] = result;
        }
    }
}

JsonConvert.DeserializeObject is from Json.NET

Link: Serializing and Deserializing JSON with Json.NET

Towards the second half of Create REST API using ASP.NET MVC that speaks both JSON and plain XML, to quote:

Now we need to accept JSON and XML payload, delivered via HTTP POST. Sometimes your client might want to upload a collection of objects in one shot for batch processing. So, they can upload objects using either JSON or XML format. There's no native support in ASP.NET MVC to automatically parse posted JSON or XML and automatically map to Action parameters. So, I wrote a filter that does it."

He then implements an action filter that maps the JSON to C# objects with code shown.

First download this JavaScript code, JSON2.js, that will help us serialize the object into a string.

In my example I'm posting the rows of a jqGrid via Ajax:

    var commissions = new Array();
    // Do several row data and do some push. In this example is just one push.
    var rowData = $(GRID_AGENTS).getRowData(ids[i]);
    commissions.push(rowData);
    $.ajax({
        type: "POST",
        traditional: true,
        url: '<%= Url.Content("~/") %>' + AREA + CONTROLLER + 'SubmitCommissions',
        async: true,
        data: JSON.stringify(commissions),
        dataType: "json",
        contentType: 'application/json; charset=utf-8',
        success: function (data) {
            if (data.Result) {
                jQuery(GRID_AGENTS).trigger('reloadGrid');
            }
            else {
                jAlert("A problem ocurred during updating", "Commissions Report");
            }
        }
    });

Now on the controller:

    [HttpPost]
    [JsonFilter(Param = "commissions", JsonDataType = typeof(List<CommissionsJs>))]
    public ActionResult SubmitCommissions(List<CommissionsJs> commissions)
    {
        var result = dosomething(commissions);
        var jsonData = new
        {
            Result = true,
            Message = "Success"
        };
        if (result < 1)
        {
            jsonData = new
            {
                Result = false,
                Message = "Problem"
            };
        }
        return Json(jsonData);
    }

Create a JsonFilter Class (thanks to JSC reference).

    public class JsonFilter : ActionFilterAttribute
    {
        public string Param { get; set; }
        public Type JsonDataType { get; set; }
        public override void OnActionExecuting(ActionExecutingContext filterContext)
        {
            if (filterContext.HttpContext.Request.ContentType.Contains("application/json"))
            {
                string inputContent;
                using (var sr = new StreamReader(filterContext.HttpContext.Request.InputStream))
                {
                    inputContent = sr.ReadToEnd();
                }
                var result = JsonConvert.DeserializeObject(inputContent, JsonDataType);
                filterContext.ActionParameters[Param] = result;
            }
        }
    }

Create another class so the filter can parse the JSON string to the actual manipulable object: This class comissionsJS are all the rows of my jqGrid.

    public class CommissionsJs
    {
        public string Amount { get; set; }

        public string CheckNumber { get; set; }

        public string Contract { get; set; }
        public string DatePayed { get; set; }
        public string DealerName { get; set; }
        public string ID { get; set; }
        public string IdAgentPayment { get; set; }
        public string Notes { get; set; }
        public string PaymentMethodName { get; set; }
        public string RowNumber { get; set; }
        public string AgentId { get; set; }
    }

I hope this example helps to illustrate how to post a complex object.

No need to do anything special. only in your post section do as follows:

    $.post(yourURL,{ '': results})(function(e){ ...}

In server use this:

   public ActionResult MethodName(List<yourViewModel> model){...}