cannot get the ViewModel of a Partial View cshtml

Question

I have a PartialView on the index page as follows :-

@{ Html.RenderPartial("ImageUpload"); }

and the PartialView looks like this :-

@model MvcCommons.ViewModels.ImageModel

<p>
   @Html.ActionLink("Create New", "Create")
</p>
<table>
    @foreach (var item in Model) {
      <tr>
        <td>
          @Html.DisplayFor(modelItem => item.FileName)
        </td>
        <td>
          @Html.DisplayFor(modelItem => item.Description)
        </td>
      </tr>
    }
</table>

Now according to the declaration in the top of the PartialView, its supposed to go inside the ViewModels.ImageModel, and inside that class I have a constructor :-

public ImageModel()
{
   XDocument imageMetaData = XDocument.Load(uploadsDir + @"/ImagesMetaData.xml");
   var images = from image in imageMetaData.Descendants("image")
                select new Image(image.Element("filename").Value,
                image.Element("description").Value);
   this.AddRange(images.ToList<Image>());
}

However, for some reason, in the ImageUpload partial View, when I debug, I am not being redirected to this ViewModel constructor, and as so the model inside the PartialView is null.

Am I missing something here?

How can I get it to actually pass through my constructor? Do I also need to do an @model in the main Index page (where is hosting the PartialView).

Thanks for your help and time

Answer 1

The line @model MvcCommons.ViewModels.ImageModel is used to declare a strongly typed Model, but not instanciate it.

You should use

@{ Html.RenderPartial("ImageUpload", <yourmodel>); }

or more simple :

@Html.Partial("ImageUpload", <yourmodel>)

By the way, in your case:

@Html.Partial("ImageUpload", new ImageModel())

But carefull: your Model should be constructor less and not loading/parsing XML. This should be done in a Controller (and set in the Caching system?).

If you wish to keep your main view Model less, you can also create an Action with the Attribute ChildActionOnly for a partial rendering, and call it with @Html.Action(...): it creates a new ControllerContext.