assigning Datasource of kendo grid in Javascript part. (Asp.net MVC-Razor Engine)

Question

I want to assign the datasource of the kendo grid in Javascript part of the View, as i want to display it in response to selecting another object from another kendogrid on the same page, so i want to pass the value of this object to the controller back and then view the grid.

I found the best way is to assign the datasource in the javascript function related to the selection part, if there is better ideas, please suggest it.

Kendo grid :

@(Html.Kendo().Grid<dynamic>()
.Name("StatusGrid")
//.HtmlAttributes(new { style="width:50%;" })
.Columns(columns =>
{
    foreach (System.Data.DataColumn c in Model.GridStatus.Columns)
    {
        columns.Bound(c.ColumnName).EditorTemplateName("String");
    }
})
.DataSource(dataSource => dataSource
    .Ajax()
    .Events(events => events.Error("error_handler"))
    .Model(model =>
    {
        foreach (System.Data.DataColumn column in Model.GridStatus.Columns)
        {
            model.Field(column.ColumnName, column.DataType);
            model.Id("Id");
        }
    })
    //.Read(read =>

    //  read.Action("ActivityGridDisplay", "Configuration")
    //)
))

So how to write this part in the javascript:( assigning the read and the model) ?

    .DataSource(dataSource => dataSource
    .Ajax()
    .Events(events => events.Error("error_handler"))
    .Model(model =>
    {
        foreach (System.Data.DataColumn column in Model.GridStatus.Columns)
        {
            model.Field(column.ColumnName, column.DataType);
            model.Id("Id");
        }
    })
    .Read(read =>

        read.Action("ActivityGridDisplay", "Configuration")
    )
))

Answer 1

It can be done in an easier way, just to assign or bind the read part using ajax, even you can pass some variables with that read part using javascript which i find easier than kendo:

function onChange(e) {

    var grid = $("#grid").data("kendoGrid");
    var dataRows = grid.items();
    var rowIndex = dataRows.index(grid.select());
    var selectedname = grid.dataItems()[rowIndex];
    document.getElementById("ActivityGrid").style.bottom = "0px";

    $.ajax({

        url: '/Configuration/ActivityGridDisplay',
        contentType: 'application/html charset=utf-8',
        type: 'GET',
        dataType: 'html',
        data: { 'nodeName': selectedname.Name, 'nodeType': selectedname.Type, 'nodeID': selectedname.NodeId },
        success: function (data) {
            $('#body').html(data);
        }
    })


}

and the edited part of the question