Returning List<Object> in HTML.BeginForm MVC 5

Question

Hey Guys I need help on this, I know I can return List<string> in HTML.BeginForm which looks like this:

@using (Html.BeginForm("Test", "Home", FormMethod.Post, new {@class = "form-horizontal", role = "form"}))
{
<textarea name="logic" style="width: 10em; height: 10em;"></textarea>
<textarea name="logic" style="width: 10em; height: 10em;"></textarea>

<div class="form-group">
    <div class="col-md-offset-2 col-md-10">
        <input type="submit" class="btn btn-default" value="Register" />
    </div>
</div>
}

I can return List<string> Logic on my Controller HTTPOST

But my question is how will I do it if I need to return a List of Object? Not using a Model because I need to return 2 Objects so I am was thinking like my controller should get it something like this

        public ActionResult Test(List<Model1> model1, List<Model2> model2)

Thanks for those who can help

Unclear what your asking. What are you models Model1 and Model2? — user3559349
– user3559349, Commented Sep 18, 2015 at 3:12
Model1 and Model2 are just examples, let say they have 2 parameters which are string1 and string2 for each Model if it will make it clearer — Franz Justin Buenaventura
– Franz Justin Buenaventura, Commented Sep 18, 2015 at 3:15
This site is for helping you with your code. Start with a view model containing 2 collection properties and then pass it to the view. Use a for loop or EditorTemplate's for you model types and post back you view model. — user3559349
– user3559349, Commented Sep 18, 2015 at 3:22
Refer this answer for an example of generating controls in a for loop and this one for using and EditorTemplate — user3559349
– user3559349, Commented Sep 18, 2015 at 4:15

Murtaza Tahir Ali · Accepted Answer · 2015-09-18 06:18:28Z

You can do in provided way

First declare class

 public class TestModel
    {
        public string Name { get; set; }
        public string Text { get; set; } 
    }

Second Controller to accept list of model

 [HttpPost]
        public ActionResult Test(List<TestModel> model1, List<TestModel> model2)
        {
            return View();
        }

Last, View to pass model values

@using (Html.BeginForm("Test", "Home", FormMethod.Post, new {@class = "form-horizontal", role = "form"}))
{
<textarea name="model1[0].Name" style="width: 10em; height: 10em;"></textarea>
<textarea name="model1[0].Text" style="width: 10em; height: 10em;"></textarea>
<textarea name="model1[1].Name" style="width: 10em; height: 10em;"></textarea>
<textarea name="model1[1].Text" style="width: 10em; height: 10em;"></textarea>
<textarea name="model2[0].Name" style="width: 10em; height: 10em;"></textarea>
<textarea name="model2[0].Text" style="width: 10em; height: 10em;"></textarea>
<textarea name="model2[1].Name" style="width: 10em; height: 10em;"></textarea>
<textarea name="model2[1].Text" style="width: 10em; height: 10em;"></textarea>

<div class="form-group">
    <div class="col-md-offset-2 col-md-10">
        <input type="submit" class="btn btn-default" value="Register" />
    </div>
</div>
}

Put the breakpoint inside Test controller you will see list populated

Yogi · Accepted Answer · 2015-09-18 04:08:50Z

3

I would suggest to wrap these lists into a view model class like this -

public class ModelView
{
    public List<Model1> model1 { get; set; }
    public List<Model2> model2 { get; set; }
}

...and then the controller will look like -

public ActionResult Test(ModelView modelView)

Within the controller you can easily access the lists.

modelView.model1 
modelView.model2

answered Sep 18, 2015 at 4:08

Yogi

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2 Comments

Franz Justin Buenaventura Over a year ago

Thats a good idea, but what will i put on the name and id of the html inputs? As much i do not like to use editorfor but in your solution i need to use @model to implement it

Yogi Over a year ago

Bind your View from this ModelView class. Now you can easily have controls for this like - model.<propertyName> OR model.<propertyName>.<subpropertyName> etc for example. Comment from Steplen gives good hint

Rashedul.Rubel · Accepted Answer · 2015-09-18 05:52:10Z

If you want to generate a view with multiple model objects then you need to create a ViewModel comprising properties that are needed from those models. And then reference the view with this ViewModel.

public class Model1
{
    public string prop11 { get; set; }
    public string prop12 { get; set; }
}

public class Model2
{
    public string prop21 { get; set; }
    public string prop22 { get; set; }
}

public class ViewModel
{
    public List<Model1> model1 { get; set; }
    public List<Model2> model2 { get; set; }
}

Then generate the view referencing the viewmodel that will get the properties from both models.

controller action that will be hit from that view:

public ActionResult Test(ModelView modelView) // you can access the viewmodel properties

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