MVC4 Pass model via Ajax.BeginForm

Question

I have tried to follow some good posts here to get this to work but every time i click the ajax postback the break point in my controller shows a model with null values. This page displays model values for viewing only, but I have tried putting them in their own form and also wrapping them in the ajax form and nothing seems to work.

@model VendorProfileIntranet.Models.VendorProfile

$(function () {
$("form").submit(function () {
    if ($(this).valid()) {
        $.ajax({
            url: this.action,
            type: this.method,
            data: $(this).serialize(),
            success: function (result) {
                $("#message").html(result);
            }
        });
    }
    return false;
});

});

    @using (Ajax.BeginForm("SelectVendor", "Home", new AjaxOptions { HttpMethod="post", InsertionMode=InsertionMode.Replace, UpdateTargetId="message" }))
{
    <div style="float:right; width:500px">
        <div id="message"></div>
        <input id="btnSubmit" type="submit" value="Select Vendor" />
    </div>

The view is really long (just displays model values for viewing) abbreviated here.

<div id="viewform">
    <div id="viewform-contact" style="float:left;">
    <fieldset>
        <legend>Contact Information</legend>
        @Html.HiddenFor(model => model.ProfileID)
        <div class="view-field">
            @Html.LabelFor(model => model.Name)
            @Html.DisplayFor(model => model.Name)
        </div>
        <br />
        <div class="view-field">
            @Html.LabelFor(model => model.Email)
            @Html.DisplayFor(model => model.Email)
        </div>
        <br />
        <div class="view-field">
            @Html.LabelFor(model => model.Phone)
            @Html.DisplayFor(model => model.Phone)
        </div>
        <br />
        <div class="view-field">
            @Html.LabelFor(model => model.Website)
            @Html.DisplayFor(model => model.Website)
        </div>
        <br />
        <div class="view-field">
            @Html.LabelFor(model => model.CompanyName)
            @Html.DisplayFor(model => model.CompanyName)
        </div>
        <br />
        <div class="view-field">
            @Html.LabelFor(model => model.Address1)
            @Html.DisplayFor(model => model.Address1)
        </div>
        <br />
        <div class="view-field">
            @Html.LabelFor(model => model.Address2)&nbsp;
            @Html.DisplayFor(model => model.Address2)
        </div>
        <br />
        <div class="view-field">
            @Html.LabelFor(model => model.City)
            @Html.DisplayFor(model => model.City)
        </div>
        <br />
        <div class="view-field">
            @Html.LabelFor(model => model.State)
            @Html.DisplayFor(model => model.State)
        </div>
        <br />
        <div class="view-field">
            @Html.LabelFor(model => model.Zip)
            @Html.DisplayFor(model => model.Zip)
        </div>
        <br />
        <div class="view-fieldwide">
            @Html.LabelFor(model => model.VendorNotes)
        </div>
        <br />
        <div class="view-fieldwide">
            @Html.DisplayFor(model => model.VendorNotes)
        </div>
        <br /><br />
        </fieldset>
    </div>
}

Controller. I have tried the above in one or two forms. With the above code the model contains all empty values except the ProfileID which is stuffed in a hidden input field. I think passing the model should work without having to create a hidden field for every model value?

[HttpPost]
        public ActionResult SelectVendor(VendorProfile pageModel)
    {
        // handle selected vendor
            _DAL.SelectVendor(pageModel.ProfileID);
            return Content("This Vendor has been selected", "text/html");
    }

ANSWER

The model values are not binded due to the use of DisplayFor

Answer 1

As noted in the comments, you're actually expecting to see items in your Model that you have put in your form using DisplayFor. DisplayFor fields will not post, they are simply for display. In order to post those values back, you'd need a HiddenFor for each of the field that you want to post, in order for model binding to work it's magic.