Show view returned by my controller

1

I have my Petition.

$.ajax({
    type: "POST",
    url: "Estimates/ManageProject",
    data: JSON.stringify({
        IdProyecto: dataItem.IdProyecto,
        Homologacion: dataItem.Homologacion,
        Cliente: dataItem.Cliente,
        TipoProyecto: dataItem.TipoProyecto,
        Contacto: dataItem.Contacto,

    }),
    contentType: "application/json",
    success: function (result) {
        window.open()
    }
});
return false;

And my controller

public ActionResult Estimates(ProjectModel model)
{
   return View(model);
}

How can I make that view appear in the browser? so that this view runs ordinary? or how can I do it in the best possible way?

    
asked by Kmiilo Berrio Montoya 15.09.2016 в 21:03
source

2 answers

2

With jquery.post you can only do POST ajax and since you want to perform a traditional POST what you should do is simulate the POST of a form. For this we create a temporary form and add the data that we want to send in the.

$('<form action="Estimates" method="POST"/>')
    .append($('<input type="hidden" name="IdProyecto" value="' + dataItem.IdProyecto+ '">'))
    .append($('<input type="hidden" name="Homologacion" value="' + dataItem.Homologacion+ '">'))
    .append($('<input type="hidden" name="Cliente" value="' + dataItem.Cliente+ '">'))
    .append($('<input type="hidden" name="TipoProyecto" value="' + dataItem.TipoProyecto+ '">'))
    .append($('<input type="hidden" name="Contacto" value="' + dataItem.Contacto+ '">'))
    .appendTo($(document.body))
    .submit();
    
answered by 15.09.2016 / 21:23
source
0

Try this:

$.ajax({
                                type: "POST",
                                url: "Estimates/ManageProject",
                                data: JSON.stringify({
                                    IdProyecto: dataItem.IdProyecto,
                                    Homologacion: dataItem.Homologacion,
                                    Cliente: dataItem.Cliente,
                                    TipoProyecto: dataItem.TipoProyecto,
                                    Contacto: dataItem.Contacto,

                                }),
                                contentType: "application/json",
                                success: function (result) {
                                   var w = window.open();
                                  $(w.document.body).html(result);
                                }
                            });
                            return false;
    
answered by 15.09.2016 в 21:22