Question

What are the ways to render the input value filename and send it to a controller :

<div id="fileuploaddiv" class="fileuploaddivclass">
<form action="@Model.FormAction" method="@Model.FormMethod"
    enctype="@Model.FormEnclosureType">
<input type="hidden" name="key" value="uploads/${filename}" id="filename" />
<input type="hidden" name="AWSAccessKeyId" value="@Model.AWSAccessKey" />
<input type="hidden" name="Content-Type" value="image/jpeg">
<div>
    Please specify a file, or a set of files:
    <input type="file" name="file" />
</div>
<input type="submit" value="Upload" />
</form>
</div>

Answer 1

You need to look up some MVC3 conventions (I’d recommend NerdDinner as a good starting tutorial), but here is a somewhat similar approach to what you want to do:

@Model YourViewModel
<div id="fileuploaddiv" class="fileuploaddivclass">
    @using(Html.BeginForm(Model.FormAction, Model.FormController, FormMethod.Post)
        @Html.HiddenFor(model.key => ${fileName})
        @Html.HiddenFor(model.AWSAccessKeyID)
        @Html.HiddenFor(model.Content-Type)
        @<input type="submit" value="Submit My Form" />
    @Html.EndForm()
</div>

Your model would look like (And I’m confused here because you seem to be dynamically setting the controller and action, which is unusual):

public class YourViewModel
{
     public string FormAction { get; set; }
     public string FormController { get; set; }
     public int AWSAccessKeyID { get; set; }
     public string Content-Type { get; set; }
}

Now on to controllers:

[HttpGet]
public ActionResult WhateverControllerName()
{
    YourViewModel yvm = new YourViewModel();
    //Initalize viewmodel here
    Return view(yvm);
}

[HttpPost]
public ActionResult WhateverControllerName(YourViewModel yvm)
{
    if (ModelState.IsValid) {
        //Do whatever you want here. Perhaps a redirect?
    }
    return View(yvm);
}

Note: I am garbage at syntax, so you’ll have to check this, but Visual Studio should tell you what works.