How to make a LINQ/lambda and consume it in the view?

Question

I have an object called Radio:

public class Radio
{
    public int RadioId { get; set; }
    public string StreamUrl { get; set; }
    public bool Ativo { get; set; }
    public bool Mp3 { get; set; }
}

I’m doing a select on controller and trying to send her to View:

    var RadioAtivo = db.Radios
        .Where(p => p.Ativo == true && p.Mp3 == false)
        .Take(1);

return View(RadioAtivo.ToList());

In the view, would like to recover, for example, the field StreamURL

@model List<RadioFM.Domain.Domain.Radio>

<p>@Html.DisplayFor(r => r.???) </p>

I thought I’d put in a ViewBag, but I found it a bit tricky. I thought of creating a ViewModel already with the data, but I thought it would be kind of duplicate a problem.

What’s the simplest solution in this case?

Answer 1

It doesn’t make sense here:

var RadioAtivo = db.Radios
    .Where(p => p.Ativo == true && p.Mp3 == false)
    .Take(1);

return View(RadioAtivo.ToList());

If you want to present only one element on screen, first of all, you do not need to use Where() with the Take(). Not that it doesn’t work, but it returns a list, and what you want is just an element.

You must use First() or Single() (if you are sure the element exists) or FirstOrDefault() if you do not have this certainty (the first two give error if there is no element):

var RadioAtivo = db.Radios
    .FirstOrDefault(p => p.Ativo == true && p.Mp3 == false);

The object passed to the return, therefore, it is only an element that does not need to be converted to ToList():

return View(RadioAtivo);

To View, in turn, it doesn’t work with a list, so it doesn’t make sense for you to use it here:

@model List<RadioFM.Domain.Domain.Radio>

@model is for you to say to the View what will be placed inside the variable Model (in capital letters). The correct one is only the class declaration, as below:

@model RadioFM.Domain.Domain.Radio

(an observation: this name of namespace RadioFM.Domain.Domain.Radio has everything to be an anti-standard and therefore a bad practice)

Accordingly, the StreamUrl can be displayed like this:

<p>@Html.DisplayFor(model => model.StreamUrl)</p>

Or still simply so, if there is no formatting configured in the project:

<p>@Model.StreamUrl</p>

Answer 2

The simplest is the one you are doing even. Others form as the ViewBag or create a view model may be useful in certain situations, but are more complex and therefore should only be used if the code requires its use to achieve the result.

I don’t think you need a dynamic way to pass the object, which is the function of ViewBag. Nor do you need an intermediary where you need to do something extra or better organize the data to be used in view:

<p>@Html.DisplayFor(r => r.StreamUrl) </p>

I would change the query expression to take the redundancy and maybe change the template to deliver an item as you wish (as comments):

var RadioAtivo = db.Radios
    .Where(p => p.Ativo && !p.Mp3)
    .Take(1);
return View(RadioAtivo.ToList()[0]);

Or better yet:

var RadioAtivo = db.Radios
    .Where(p => p.Ativo && !p.Mp3)
    .First();
return View(RadioAtivo.ToList());

As you will receive only one item you will not receive a list, so change:

@model RadioFM.Domain.Domain.Radio

I put in the Github for future reference.

And confirm what I said in the comments, rethink this RadioFM.Domain.Domain.Radio, this doesn’t seem to be necessary.