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Home/ Questions/Q 7030539
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T00:41:02+00:00 2026-05-28T00:41:02+00:00

I have the following view: @model ViewModels.Shared.BaseViewModel <div class=rep_tb0> <div class=rep_tr0> <div class=rep_td0>@Html.LabelFor(model =>

  • 0

I have the following view:

    @model ViewModels.Shared.BaseViewModel
    <div class="rep_tb0">
        <div class="rep_tr0">
            <div class="rep_td0">@Html.LabelFor(model => Model.Link.Position)</div>
            <div class="rep_td0">@Html.LabelFor(model => Model.Link.Title)</div> 
        </div>
        @for (var i = 0; i < Model.AdminDetails.Count; i++)
        {
            var qs = "&ac=" + Model.Meta.AccountID +
                     "&me=" + Model.AdminDetails[i].RowKey;  
        <div class="rep_tr0">
            <div class="rep_td0">
                @Html.TextBox(("Position_" + @i), Model.AdminDetails[i].Position, new { size = 2 })
            </div>
            <div class="rep_td0">@Model.AdminDetails[i].Title</div>             
        </div>  
        }
    </div>    

Populated by this controller action:

    public ActionResult Detail(string ac, string me)
    {
            vm.AdminDetails = _link.Detail(ac + me).ToList();
        return View(vm);
    }

I would like to use Ajax, jQuery and HTML.Renderpartial as there are times when I need to request a refresh of the data for example when I change the value of position.

I am totally new to the idea of doing this. I found links that help me but mostly it was about Ajax with earlier versions of MVC and not so much about MVC3. Can someone give me some pointers. I am not looking for code. Just a few line example of how I could go about doing the coding in my action method, how I should call the Ajax etc.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-28T00:41:03+00:00Added an answer on May 28, 2026 at 12:41 am

    You need a child action, not a partial view.

    You can then write

    <div id="#thingy">@Html.Action("Detail", new { ac, me })</div>
    
    $('#thingy').load("@Server.JavaScriptStringEncode(Url.Action("Detail", new { ac, me }))");
    
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