ASP.NET MVC & C# : pass a viewmodel to controller

Question

I have a controller that passes the viewmodel to a view, and in the view it creates the table correctly.

Then in the same page of the table I have a link that redirects the user to another page.

This new page needs to keep the viewmodel of the previous view. How could I do it?

In the view, in the table page this link that should redirect to an action of EanMatch controller:

@Html.ActionLink("Inserimento manuale", "Index", "EanMatch",  null, new {Vm = Model})

The model is this:

public List<DDTViewModel> DDTList { get; set; }

In the EanMatchController and Index action I have:

[HttpGet]
public ActionResult Index(DDTListViewModel Vm)
{
    ....
}

I can't understand why it doesn't work. DDTListViewModel Vm in Index actions takes 'null' value, meanwhile I have seen with debug that Model got the list of data.

How can I pass the viewmodel to the action argument correctly?

Any doubt, ask !

What is DDTListViewModel ? Is it smth that safety store on the client-side? — Basil Kosovan
– Basil Kosovan, Commented Dec 5, 2019 at 16:11
@staff614 you are setting the ViewModel inside a property inside an anonymous object. This is why you are not getting the result you want. — panoskarajohn
– panoskarajohn, Commented Dec 5, 2019 at 16:12
@BasilKosovan DDTListViewModel contains the list of a viewModel public List<DDTViewModel> DDTList { get; set; } — staff614
– staff614, Commented Dec 5, 2019 at 16:22

Rahul Sharma · Accepted Answer · 2019-12-06 13:16:23Z

1

For your particular case, You'll have to serialize your model as a JSON string, and send that to your controller to turn into an object.

Here's your ActionLink:

@Html.ActionLink("Inserimento manuale", "Index", "EanMatch",  new { jsonModel= Json.Encode(Model.DDTListViewModel) }, null)

And your Controller method would look like:

public ActionResult Index(string jsonModel)
{
  var serializer= new DataContractJsonSerializer(typeof(Model.DDTListViewModel));
  var yourmodel=  (DDTListViewModel)serializer.ReadObject(GenerateStreamFromString(jsonModel));
}

answered Dec 6, 2019 at 13:16

Rahul Sharma

8,3772 gold badges39 silver badges58 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

panoskarajohn · Accepted Answer · 2019-12-05 16:20:42Z

1

@Html.ActionLink("LinkName", "Index", "EanMatchController", new ViewModel { Id = 1, Name = "Sth", Description = "Hello" }, null);

You set the object as property to an anonymous object. This is why your object is not passed on the controller.

It is like having a wrapper object around your object like below.

public class Sth{
 public ViewModel { get; set; }
}

So C# does not recognize this as your object you need to somehow create it inside razor and send it back to the controller. The code above works. Try a simple example.

edited Dec 5, 2019 at 16:20

answered Dec 5, 2019 at 16:09

panoskarajohn

2,0082 gold badges25 silver badges41 bronze badges

2 Comments

staff614 Over a year ago

it gives error to 'new viewModel'. however, i tried to put new { id = 1 } and it passes the data to the action, it just doesn't work with the list of viewmodel

panoskarajohn Over a year ago

Pass your own view model. I just posted as an example. View model is an object with 3 properties.

terrencep · Accepted Answer · 2019-12-05 16:04:23Z

0

if you controller name is EanMatch, then the next issue is that I think you are passing the model in where it expects htmlattributes. So instead of null, new {Vm = Model}; switch them.

@Html.ActionLink("Inserimento manuale", "Index", "EanMatch", new {Vm = Model} ,null )

answered Dec 5, 2019 at 16:04

terrencep

6735 silver badges16 bronze badges

2 Comments

terrencep Over a year ago

Hit f12 and see what's being send post/get.

staff614 Over a year ago

it posts as query string ?Capacity=16&Count=11 which I can't explain actually

Johnathan Barclay · Accepted Answer · 2019-12-05 16:07:26Z

0

You don't need to create the anonymous object, just pass your view model:

@Html.ActionLink("Inserimento manuale", "Index", "EanMatch", null, Model)

The framework will convert your model to a query string that will be bound by your controller.

answered Dec 5, 2019 at 16:07

Johnathan Barclay

20.5k2 gold badges27 silver badges40 bronze badges

Comments

Basil Kosovan · Accepted Answer · 2019-12-05 16:12:33Z

0

Try this:

@Html.ActionLink("Inserimento manuale", "Index", "EanMatch",  null,  Model)

OR: Each request to the server has its own context(thread) and doesn't know anything about others. If you need to share a model or another data between a few different requests to the server you can use temp data or session. Inside of the first action, you need to save this model:

Session["myModel"] = Model;

and read in another:

var val = Session["myModel"];

edited Dec 5, 2019 at 16:12

answered Dec 5, 2019 at 16:07

Basil Kosovan

1,0991 gold badge12 silver badges37 bronze badges

1 Comment

staff614 Over a year ago

Where actually should "Session["myModel"] = Model;" go? it is between a View to Controller, not action to action

Collectives™ on Stack Overflow

ASP.NET MVC & C# : pass a viewmodel to controller

5 Answers 5

Comments

2 Comments

2 Comments

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

2 Comments

2 Comments

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related