[Solved] Telerik MVC 3 Grid multilevel (3 level) master-detail

[Samantha]

Here's my grid

Class year > Course > Student

Problem 1. How can I get the key from the first grid? (classyearID from classyear grid?)

The reason I need it because I have more than 5 class years, but only have 2 courses. so basically those two coursse will exist in every class year. and since I cant get the class yearID, students are reappearing on each class year, please help

check my view below:

@(Html.Telerik().Grid<ClassYear>().HtmlAttributes(new { style = "width: 100%" })
.Name("grdClassYear")
.DataBinding(binding => binding.Ajax()
    .Select("GetClassYears", "Home"))
.DataKeys(keys => keys
    .Add(o => o.ClassYearID)
    .RouteKey("classyearID"))
.Columns(cols =>
{
    cols.Bound(c => c.ClassYearDate);
    cols.Bound(c => c.Name);
})

.DetailView(course => course.ClientTemplate(
            Html.Telerik().Grid<Course>()
            .Name("grdCourse_<#= ClassYearID #>")
            .DataBinding(dataBinding => dataBinding.Ajax()
                .Select("GetCourses", "Home"))
            .DataKeys(keys => keys.Add(o => o.CourseID))
            .Columns(cols =>
            {
                cols.Bound(c => c.CourseName);
                cols.Bound(c => c.Description);
            })

                    .DetailView(stu => stu.ClientTemplate(
                        Html.Telerik().Grid<Student>()
                        .Name("grdStudent_<#= CourseID #>")
                        .DataBinding(dataBinding => dataBinding.Ajax()
                            .Select("GetStudentsA", "Home", new { classyearID = "<#= ClassYearID #>", courseID = "<#= CourseID #>" })))
                        .DataKeys(keys => keys.Add(o => o.PersonID))
                        .Columns(cols =>
                        {                                
                            cols.Bound(c => c.PersonID).ReadOnly().Hidden();
                            cols.Bound(c => c.MidshipmenNumber);
                            cols.Bound(c => c.LastName);
                            cols.Bound(c => c.FirstName);
                        })

                    .Sortable()
                    .ToHtmlString()
                    ))