Thursday, April 02, 2009

Json for jqGrid from ASP.Net MVC

jqGrid takes a specific format for its json (taken from jqGrid documentation):

total: "xxx",page: "yyy", records: "zzz",
rows : [
{id:"1", cell:["cell11", "cell12", "cell13"]},
{id:"2", cell:["cell21", "cell22", "cell23"]},

The tags mean the following:

total - Total number of Pages.
page - Current page Index.
records - Total number of records in the rows group.
rows - An array with the data plus an identifier.
id - The unique row identifier, needs to be an int from what I have found.
cell - An array of the data for the grid.

The ASP.Net MVC framework has the JsonResult response type which we can use to populate the jqGrid. As an example I created a Person model and a method to return some data:

public class Person
public int ID { get; set; }
public string Name { get; set; }
public DateTime Birthday { get; set; }

public IEnumerable<Person> GetABunchOfPeople()
yield return new Person {ID = 1, Name = "Darren", Birthday = new DateTime(1970, 9, 13)};
yield return new Person {ID = 2, Name = "Dawn", Birthday = new DateTime(1971, 6, 1)};
yield return new Person {ID = 3, Name = "Thomas", Birthday = new DateTime(1995, 10, 3)};
yield return new Person {ID = 4, Name = "Zoey", Birthday = new DateTime(1997, 8, 15)};

Generating the JSON is as follows, this going into a PersonModel class:

public JsonResult GetABunchOfPeopleAsJson()
var rows = (GetABunchOfPeople()
.Select(c => new
id = c.ID,
cell = new[]
return new JsonResult
Data = new
page = 1,
records = rows.Length,
total = 1

The controller then would look like:
public class PersonController : Controller
public ActionResult Index()
return View();
public JsonResult GetAllPeople()
var model = new Models.PersonModel();
return model.GetABunchOfPeopleAsJson();

The view doesn't need to inherit any model other than View page if all you want is for the jqGrid to show the data. In the view, Index.aspx in this case, you add table and div for the jqGrid:

<table id="dictionary" class="scroll" cellpadding="0" cellspacing="0"></table>
<div id="pager" class="scroll" style="text-align: center;"></div>

Configuring jqGrid is:
[js highlight="6,8"]
$(document).ready(function() {

caption: "Tank Dictionary",
pager: $("#pager"),
url: '<%= ResolveUrl("~/Person/GetAllPeople") %>',
editurl: '&lt;%= ResolveUrl("~/Person/Edit") %&gt;',
datatype: 'json',
myType: 'GET',
colNames: ['ID', 'Name', 'Birthday'],
colModel: [
{ name: 'ID', index: 'ID', width: 150, resizable: true, editable: false },
{ name: 'Name', index: 'Name', width: 200, resizable: true, editable: true },
{ name: 'Birthday', index: 'Birthday', width: 300, resizable: true, editable: true }
sortname: 'ID',
sortorder: 'desc',
viewrecords: true,
height: '100%',
imgpath: '<%= ResolveUrl("~/Scripts/jquery/jqGrid/themes/basic/images") >'
The url tag is set to call the Person controller to get the JSON results. Note the datatype is json.

The editurl tag will be talked about later.


Vinoth said...

Check out this, might be helpful

Rahuman said...

Similar article with fully functional example at

raghava said...

Its Great working with jqgid,but i need manipulations with data and with all features in jqgrid such as edit,subgrid,delete,select,update and ssupported js files for above features