How to send model object in Html.RenderAction (MVC3)

Question

I'm using MVC3 razor, and I'm trying to pass an object to a partial view, and it's not working.

This works fine without sending the object model to the partial view:

Html.RenderAction("Index", "ViewName");

Trying this doesn't sent the model object, i'm getting nulls instead (the object has data, and the view expects it):'

Html.RenderAction("Index", "ViewName", objectModel);

Is this even possible using RenderAction?

Thanks!

Edit: I found the error, there was an error with the controller's action that didn't pick up the sent object. Thanks for all your help!

could you provide the signature of the action you are calling? — Ivo, Jan 22 '12 at 06:26
why are you passing the viewname as a parameter? you should be passing the action and the controller instead — Ivo, Jan 22 '12 at 06:27
@Html.RenderAction("Index", "ViewName", new {objectParam1 = objectModel1, objectParam1 = objectModel2}) — Mangesh Pimpalkar, Jan 22 '12 at 06:30

score 35 · Answer 1 · edited Apr 27 '18 at 11:27

You can actually pass an object to a controller method using Action. This can be done on any avaialble view, for instance I have one in a shared library that gets built to project bin folders that reference my shared project (properties - Copy if newer on the view file, in Visual Studio). It is done like so:

Controller:

public class GroovyController : Controller
{
    public ActionResult MyTestView(MyModel m)
    {
        var viewPath = @"~\bin\CommonViews\MyTestView";
        return View(viewPath, m);
    }
}

MVC page (using Razor syntax):

@Html.Action("MyTestView", "Groovy", new { m = Model })

or using RenderAction method:

@{ Html.RenderAction("MyTestAction", "MyTestController", new { area = "area", m = Model }); }

Note: in the @Html.Action(), the Model object must be of type MyModel and that 3rd parameter must be set to the controller variable name, of which mine is MyModel m. The m is what you must assign to, so I do m = Model.

This should be the accepted answer. Thanks for adding it a year after. — Valamas, Mar 24 '14 at 02:10

score 27 · Accepted Answer · edited May 07 '13 at 10:06

27

say you want to pass foo as model, make it first

public class Foo {
    public string Name { get; set; }
    public int Age { get; set; }
}

now make an ActionResult

public ActionResult FooBar(Foo _foo){
    return PartialView(_foo);
}

call it

@Html.RenderAction("FooBar", "Controller", new { Name = "John", Age=20 });

edited May 07 '13 at 10:06

Community

1
1

answered Jan 22 '12 at 06:34

John x

3,911
8
39
66

Thanks for your answer. I had an error that I overlooked in the Controller's Action signature. Fixed it and now it's getting the object. – Michael Jan 22 '12 at 06:54
Does this pass ModelState? – aelstonjones Dec 20 '12 at 22:34
Thanks! But I had to surround Html.RenderAction call with curly braces to make it work. Otherwise, Razor says "Cannot implicitly convert type 'void' to 'object'" – burkay Dec 04 '19 at 16:26

score 6 · Answer 3 · answered Feb 06 '13 at 11:07

Usually if I have a model already available it makes more sense to use Html.Partial than trying to render an action.

@Html.Partial("Foo", Model.FooModel)

Where Foo.cshtml is a view file (perhaps in your Shared folder) strongly typed with with @model FooProject.Models.FooModel or whatever your model is called. This can be as complex a model as you need it to be. Model is your page's main model into which you must set FooModel - or just omit this parameter if the Foo view uses the same model as the parent page.

RenderAction is generally better when you have just simple parameters, because you're just simulating a request to a regular action which has routing/query string parameters - and then dumping that response into your page. It works well if you need to put something in a Layout that isn't available in your page's model such as an element in a side bar.

How to send model object in Html.RenderAction (MVC3)

3 Answers3

Linked

Related