JQGrid Loading lots of data - asp.net-mvc

Question

SITUATION
I am using Trirand JQGrid for MVC[server side] in my proj.
I've got more than 5 hundred thousand records in a single table.
I load the data by calling this piece of code. this is what gives 500000 records collection.
IEnumerable<myIndexViewModel> myviewmodel= _allincidents.Select(x => new myIndexViewModel
{
IncidentRequestStatus = x.RequestStatus,
RequestByUserName = x.RequestByUserName,
Subject = x.Subject
});
gridModel.JqGrid.DataBind(myviewmodel.AsQueryable());
JQgrid handles the json based ajax requests very nicely for every next page i click.
PROBLEM
I dont want to load 5 hundred thousand records all together on the page load event as it kills jqgrid.
If i write a stored procedure in the DB for requesting a specific page to be displayed then its gonna load only that page in the myviewmodel collection.
How do i get pages on the fly from the DB when the next page is clicked. is this even possible in jqgrid?
SITUATION 2
Based on the answers from VIJAY and MARK the approach they have shown is absolutely correct but over here the JQGRID for MVC sets up the DATAURL property for making the method call. In this case its the IncidentGridRequest.
How do i send in the page number when the grid next page or previous page is clicked?
incidentModel.IncidentGrid.DataUrl = Url.Action("IncidentGridRequest")
public JsonResult IncidentGridRequest()
{
}

Answer 1

Your controller action that will provide your grid with results can accept some extra information from jqGrid.
public ActionResult GetGridData(string sidx, string sord, int page, int rows, bool _search, string filters)
The main parts you are interested in is the page, rows (sidx is for column sorting, sord for the sorting order, _search if there was a search done on the grid, and if so filters contains the search information)
When you generate your results you should be able to then
IEnumerable<myIndexViewModel> myviewmodel = allincidents.Select(x => new myIndexViewModel
{
IncidentRequestStatus = x.RequestStatus,
RequestByUserName = x.RequestByUserName,
Subject = x.Subject
}).Skip((page - 1) * rows).Take(rows)
PS. I'm not sure if you using IEnumberable will be moving a large amount of data from your DB but you might want to use IQueryable when you generate this subset of data for the jqGrid.
Edit: To deal with your paging issues, You should be calculating the number of total records in your query and passing that value to the grid, Ex
int totalRecords = myviewmodel.Count();
and then later you would pass that to your grid as a jSon value. Ex
var jsonData = new
{
total = (totalRecords + rows - 1) / rows,
page = page,
records = totalRecords,
userdata = new {SearchResultsFound = searchResultsFound},
rows = (
......

Answer 2

Yes, for example if you are accepting the page number you want to turn to in a variable named page and the have the size of page in a variable pageSize then:
IEnumerable<myIndexViewModel> myviewmodel = allincidents.Select(x => new myIndexViewModel
{
IncidentRequestStatus = x.RequestStatus,
RequestByUserName = x.RequestByUserName,
Subject = x.Subject
}).Skip((page-1)*pageSize).Take(pageSize));
will give you the records of size pageSize to you.

Answer 3

The Trirand jqGrid for ASP.NET MVC is using IQueryable interface inside the JqGrid.DataBind() method to implement pagin, sorting and filtering.
So the key here is to use datasource, which handle these types of operations at the database level (by crafting SQL queries to the database in such a way that only the data required is fetched). All major ORMs have this support, this includes: LINQ-2-SQL, Entity Framework, NHbiernate, LLBLGen.
You just need to use one of this technologies, and past the required context directly to JqGrid.DataBind() method (without extracting the data manually like you do it in your sample).

Answer 4

An easier approach by using PagedList library (from Nuget). There is a useful blog by Joseph Schrag
public JsonResult Users(int PageNo, int Rows)
{
var UserList = db.Users.Select(t => new
{
t.UserId,
t.Username,
t.Firstname,
t.Lastname,
t.Designation,
t.Country,
t.Email
}).OrderBy(t => t.UserId);
var pagedUserList = UserList.ToPagedList(PageNo, Rows);
var results = new
{
total = pagedUserList.PageCount, //number of pages
page = pagedUserList.PageNumber, //current page
records = UserList.Count(), //total items
rows = pagedUserList
};
return new JsonResult() { Data = results, JsonRequestBehavior = JsonRequestBehavior.AllowGet };
}