Asp.net MVC & AJAX

Yes. Its possible. Just set Application["counter"] = 0 in Application_Start function then make value increments by 1 in result view and use it to get next record.

    public PartialViewResult BtnNext()
    {
        List<Queue> model = db.Queues.OrderBy(x => x.QueueNumber).Skip(Application["counter"]).Take(1).ToList();
        Application["counter"] = Application["counter"] + 1;
        return PartialView("_queuenumber", model);
    }  

Reference Use FormCollection try following code.

public PartialViewResult BtnNext(FormCollection Form)
{
    Int32? Count = Convert.ToInt32(Form["Count"]??0);
    List<Queue> model = db.Queues.OrderBy(x => x.QueueNumber).ToList();
    model.ElementAt(count); //   [NotMapped]  public Int32? count { get; set; } add in model class
    model.count=count+1;
    return PartialView("_queuenumber", model);
} 

on view

<input type="submit" class="btn btn-primary" value="Save" id="BtnNext">
<input type="hidden" id="Count" name="Count" value="@Model.Count" />

A good practice when you realize your Views need to handle and manipulate your data, is to create a ViewModel class that wraps all the objects that you need to send to that view. In your case, you can start with a simple

public class QueueViewModel
{
    public Queue Queue { get; set ; }
    public int CurrentRecord { get; set ; }
}

Now, all you have to do is changing the action method the controller so that you initialize and pass the ViewModel to the View. It will also be better to have an optional argument acting as the default record, and then using the linq instruction Skip to go to and take a specific record:

Public PartialViewResult NextRecord(int current = 0)
{
    QueueViewModel model = new QueueViewModel();
    model.CurrentRecord = current;
    model.Queue = db.OrderBy(x => yourClause).Skip(current).Take(1);
    return PartialView(“yourview”, model);
}

I changed the List<Queue> within your model as I think you don’t need a list if you’re only showing one record at a time, but you can easily go back to the generics if you feel you really need to.

As for the view part where you handle the index on the model, there are many ways to achieve the same result. What I personally like to do is using the model to fill a data attribute of a DOM element and use that in the Ajax call. Since you now have

@model yourModelNamespace.QueueViewModel

it is possible for you to set an element (let’s say a button) to host the current value:

<button data-current-record=“@Model.CurrentRecord”>...</button>

You can now very easily retrieve that value within your Ajax call to the action method:

var currentRecord = parseInt($(‘button’).data()[currentRecord]);
$.ajax({
    url: yourPathToTheAction,
    type: ‘GET’,
    data: {
        current: currentRecord + 1
    }
});

This way you can go further and add other functions calling the same controller to move to previous record or jump to the last or the first and so on...