Web Paging with C#

Question

I’m building an administrative application, and in this application I have a message page. I am currently bringing the last 20 messages.

I already have the screen ready to receive this messages, but my problem is in making the logic of the button that show the next 20 messages, and so consecutively.

//Mensagens Enviadas
        List<Mensagem> mensagens = this.mensagemServico.GetMany(l => l.LojaId == loja.LojaId && l.OrigemId != 0).OrderByDescending(l => l.DataEnvio).Take(20).ToList();
        mensagemModel.Mensagens = mensagens;

        return View(mensagemModel);

Basically this is my routine: it takes the last 20 messages sorted by date, and according to origin (Client Origin == 0)

What I’d like to do is when I click "Next" he brings the next 20.

I had the initial idea to store in a variable of model (I have a MensagemModel to return the value to the View) the amount already brought from Messages, or current page to be able to have this control.

I’m a beginner so I have doubts how to pass some information between Views, as an example, Models, and how a pagination should be made.

If anyone can give me a hand, even if it is about the logic that should be followed to make this paging, sorry to ask this kind of question but I do not want this paging to get bad so I wanted a help from you, thank you.

Answer 1

Another good option is to use one of the two components below:

• Pagedlist.Mvc, seemingly discontinued, with support to async in this other package;
• X.PagedList.Mvc, natural continuation of Pagedlist, but supporting Portable, with support to async in this other package.

Use:

// Suponha int? pagina = null como argumento da Action.
var paginaNaoNula = pagina ?? 1;
List<Mensagem> mensagens = this.mensagemServico.GetMany(l => l.LojaId == loja.LojaId && l.OrigemId != 0).OrderByDescending(l => l.DataEnvio).ToPagedList(paginaNaoNula, 20);
mensagemModel.Mensagens = mensagens;

return View(mensagemModel);

Answer 2

You have to use more method o Skip

//Mensagens Enviadas
List<Mensagem> mensagens = this.mensagemServico.GetMany(l => l.LojaId == loja.LojaId && l.OrigemId != 0).OrderByDescending(l => l.DataEnvio)
.Skip(numberOfObjectsPerPage * pageNumber)
.Take(20).ToList();

mensagemModel.Mensagens = mensagens;

return View(mensagemModel);

The following code example demonstrates how to use Skip to ignore a specified number of elements in a sorted array and returns the remaining elements.

int[] grades = { 59, 82, 70, 56, 92, 98, 85 };

IEnumerable<int> lowerGrades =
    grades.OrderByDescending(g => g).Skip(3);

Console.WriteLine("All grades except the top three are:");
foreach (int grade in lowerGrades)
{
    Console.WriteLine(grade);
}

/*
 This code produces the following output:

 All grades except the top three are:
 82
 70
 59
 56
*/