.NET MVC3/Holding temp model

Question

I have a situation where i have to take input(form) from user. After continue button is pressed next view page is displayed. But after continue is pressed i don't want to store the model in the DB. I have to display some details(combining some tables) according to input given by the user earlier and again get some data from user. Only then i want to store the model in the respective tables. How can i perform this? I tried getting Model from user and passing to the function that generates next page. Is this is way to do it? or there is other way around?

score 1 · Accepted Answer · answered Jun 26 '12 at 09:25

Store the model submitted by the first form in session.

[HttpPost]
public ActionResult ContinueForm1(Model1 model1)
{
    if(ModelState.IsValid)
    {
       Session["Model1"] = model1;

       return View("Form2");
    }

    return View();
}

[HttpPost]
public ActionResult ContinueForm2(Model2 model2)
{
     if(ModelState.IsValid)
     {
        ... model2 is already here, get the model1 from session
        ... and save to datatbase finally return a different view or redirect to some
        ... other action
     }

     return View();
}

score 0 · Answer 2 · edited May 23 '17 at 12:04

You are heading down the right track.

You need to grab the model that is passed back from the first view - preferably you are using ViewModels here rather than binding directly to your db models. Have a look at http://lostechies.com/jimmybogard/2009/06/30/how-we-do-mvc-view-models/ and Why should I use view models? as to why these are good things.

The easiest way to do this is to pass the model in as an argument to your method e.g.

Assuming that your views are using the same ViewModel ( which may or may not be true) then you can send the viewmodel straight to your new view - else you can copy the elements into a new viewModel and send that.

e.g.

[HttpPost]
public ViewResult Step1(MyViewModel viewModel)
{
    //Do some validation here perhaps
    MySecondViewModel secondViewModel = new MySecondViewModel{
                        Id = viewModel.Id,
                        // etc. etc.
                };  
return View("Step2", secondViewModel);
}

Then you can carry on as you need until you have to persist the entity to the database. NB as you do not need to do anything special in the form to make it post the model as an argument as long as the view is strongly typed to that ViewModel.

.NET MVC3/Holding temp model

2 Answers2