Question

I have a PartialView that is an image upload, and basically I am displaying some images and then the normal Upload buttons :-

@model MvcCommons.ViewModels.ImageModel

<table>
    @if (Model != null)
    {
        foreach (var item in Model)
        {
            <tr>
                <td>
                    <img src= "@Url.Content("/Uploads/" + item.FileName)" />
                </td>
                <td>
                    @Html.DisplayFor(modelItem => item.Description)
                </td>
            </tr>    
        }
    }

</table>

@using (Html.BeginForm("Save", "File", FormMethod.Post, new { enctype = "multipart/form-data" })) {
    <input type="file" name="file" />
    <input type="submit" value="submit" /> <br />
    <input type="text" name="description" /> 
}

Now my idea is to have this in different pages. I tried it in 1 page already and is working fine, however when I Upload an image,

public ActionResult ImageUpload()
{
    ImageModel model = new ImageModel();
    model.Populate();
    return View(model);
}

I want to go back to the “previous” View, ie the View that is hosting this partial view? When I do return View(model) as above, I get into the ImageUpload partial view which I do not want to.

Thanks for your help and time.

***UPDATE*********
I went for the simple route for the time being, and hard coded the actual View name

public ActionResult ImageUpload()
{
    ImageModel model = new ImageModel(); 
    model.Populate(); 
    return View("~/Views/Project/Create.cshtml", model); 
}

however I got an error :-

The model item passed into the dictionary is of type MvcCommons.ViewModels.ImageModel, but this dictionary requires a model item of type MvcCommons.Models.Project.

Answer 1

Use the overload that takes a string of the name of the view you want.

http://msdn.microsoft.com/en-us/library/dd460310

protected internal ViewResult View(
        string viewName,
        Object model
)

i.e.

return View("ViewName", model);

if you have this in different pages then you can inject context via the action paramaters;

public ActionResult ImageUpload(string parentViewName)
{
    ImageModel model = new ImageModel();
    model.Populate();
    return View(parentViewName, model);
}

NOTE: You should only need to pass the views name not the path:

return View("Create", model);