How do i show a list from a model to another View in ASP.NET MVC 4

Question

I have a list of Topic's. Each Topic have a list of Message. When i click on "Details" I want to open a new View that display a list of the Message's from the Topic.

I expect that I need to modify this line: (this is what I have so far)

@Html.ActionLink("Details", "Details", new { id=item.Id }, null)

The Message's are accessed by item.Message.

And the first line in the Detail View

@model IEnumerable<Models.TopicMessage>

This is my Topic View:

@model IEnumerable<Models.Topic>

<table>
    <tr>
        <th>
            @Html.DisplayNameFor(model => model.Name)
        </th>
        <th></th>
    </tr>

@foreach (var item in Model) {
    <tr>
        <td>
            @Html.DisplayFor(modelItem => item.Name)
        </td>
        <td>
            @Html.ActionLink("Details", "Details", new { id=item.Id }, null)
        </td>
    </tr>
}

</table>

I can't understand. You are creating links for each topic in that line. Do you want to display a partial view for Message list? — Ufuk Hacıoğulları
– Ufuk Hacıoğulları, Commented Sep 20, 2013 at 21:03
@UfukHacıoğulları I want to open a new View with all the Message this Topic contains. — radbyx
– radbyx, Commented Sep 20, 2013 at 21:04
I have only know MVC 4 for two days. I know partial Views is for views you can reuse in other views. But I thought here the first time I should just get it to work in a normal View before advange to Partial View, but that's maybe a bad strategy? — radbyx
– radbyx, Commented Sep 20, 2013 at 21:16

p.s.w.g · Accepted Answer · 2013-09-20 21:28:49Z

1

There are a number of ways to do that. You can use the Action method to execute the controller action directly:

<tr>
    <td>
        @Html.DisplayFor(modelItem => item.Name)
    </td>
    <td>
        @Html.Action("Details", "Details", new { id=item.Id })
    </td>
</tr>

Or you can bypass the controller and render the view directly using this overload of DisplayFor:

<tr>
    <td>
        @Html.DisplayFor(modelItem => item.Name)
    </td>
    <td>
        @Html.DisplayFor(modelItem => item.Messages, "MessagesDetail")
    </td>
</tr>

Or if you define the template as the default DisplayTemplate, you can just do this:

<tr>
    <td>
        @Html.DisplayFor(modelItem => item.Name)
    </td>
    <td>
        @Html.DisplayFor(modelItem => item.Messages)
    </td>
</tr>

edited Sep 20, 2013 at 21:28

answered Sep 20, 2013 at 21:03

p.s.w.g

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8 Comments

radbyx Over a year ago

Okay I will give it a try. It's a good idea to make the View strongly typed right? It should be like this right? @model IEnumerable<Models.TopicMessage>

p.s.w.g Over a year ago

@radbyx Yes, that's definitely preferred.

radbyx Over a year ago

What about this instead @model IEnumerable<Models.Topic>? Im confused. Should I bring the Topic, that i could get the Messages from, or do I bring a list of Message?

p.s.w.g Over a year ago

@radbyx For a detail view of a single Topic entity, I'd generally recommend @model Model.Topic. This way you can enumerate the TopicMessages in that model or inspect other properties as well. If, however, this is just a list of messages (not exactuly a full detail view of the parent Topic), @model IEnumerable<Models.Topic> would be better, IMO. Of course, you might have a detail view which contains the list of messages, in which case you might have actually two views, but that's sort of up to your personal preferences.

p.s.w.g Over a year ago

@radbyx Sorry, the fourth parameter is for providing html attributes for the <a> tag. This isn't used in @Html.Action. I've updated my answer.

|

radbyx · Accepted Answer · 2013-10-18 08:46:06Z

0

I fixed this by making a somethingViewModel that contained both the liste and the Id. That way i could contain the two information from two difference places in one new object I had access to in the View.

answered Oct 18, 2013 at 8:46

radbyx

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