Open View Page with layout using jquery ajax

Question

how do I open a new page with layout using jquery ajax? I need to return strName to my view in my controller.

My jquery ajax:

mvcJqGrid.demo.edit = function (id) {
     var urlEdit = '@Url.Action("Edit")';
        $.ajax({
        type:"GET",
        url:urlEdit,
        data:{strName: $('#customerGrid').jqGrid('getCell',id,'Client00130012')}
        });
    }

Edit: *my View Controller:*

public ActionResult Edit(string strName)
    {

        var q = from c in db.CanaClie0012
                join w in db.Clientes0013 on c.Client00130012 equals w.Client0013
                where c.Client00130012 == strName
                select new ClientModel
                {
                    CanaClie0012 = new CanaClie0012()
                    {
                        Client00130012 = c.Client00130012,
                        F1Pais00200012 = c.F1Pais00200012,
                        F1Cana02530012 = c.F1Cana02530012,
                        Direcc0012 = c.Direcc0012
                    },
                    Clientes0013 = new Clientes0013()
                    {
                        Client0013 = w.Client0013,
                        Nombre0013 = w.Nombre0013,
                        F1Pais00200013 = w.F1Pais00200013
                    }
                };

        return View(q);
    }

Answer 1

You doing it in a wrong way; If you want to open edit page with your model try next.

first you need build url link in your grid to open this edit page with Model.Id. In jqGrid you need use the column formater. After that you can click on link and open your edit page like 'site.com/controller/edit/6666'

colModel: [
{ name: 'ColumnName',
    formatter: function (cellvalue, options, rowObject) {
        return '<a href="/YourController/Edit/' + cellvalue + '">' + "Edit" + '</a>';
    } 
},

],

This should work.