GridView in MVC3

When ever we need to create a grid in MVC 3, we often end up with Telerik grids, due to awesome features and support. But you need to have a licensed version of controls.
What if you don't have license for it and you have to create a Gridview of your own, like below.

Its very simple, and all you need to create is a little Html, JQuery and supporting Action Methods.
Lets see how . . .
Step 1: Create a MVC 3 Project. Create a Model class for above data.

namespace MVC3SimpleGrid.Models
{
    public class EmployeeModel
    {
        public int EmpId { get; set; }
        public string EmployeeName { get; set; }
        public string EmployeeDep { get; set; }
    }
}

Step 2: Create the Action method and the supporting functionality.
HomeController.cs:

namespace MVC3SimpleGrid.Controllers
{
    public class HomeController : Controller
    {
        static List<EmployeeModel> _lstEmployee = new List<EmployeeModel>();
        public ActionResult Index()
        {
            _lstEmployee=GetEmployees();

            return View(_lstEmployee);
        }

        private List<EmployeeModel> GetEmployees()
        {
            List<EmployeeModel> _pvtList=new List<EmployeeModel>();
            EmployeeModel mod1 = new EmployeeModel { EmpId = 1, EmployeeName = "Employee1", EmployeeDep = "EmployeeDep1" };
            EmployeeModel mod2 = new EmployeeModel { EmpId = 2, EmployeeName = "Employee2", EmployeeDep = "EmployeeDep2" };
            EmployeeModel mod3 = new EmployeeModel { EmpId = 3, EmployeeName = "Employee3", EmployeeDep = "EmployeeDep3" };
            EmployeeModel mod4 = new EmployeeModel { EmpId = 4, EmployeeName = "Employee4", EmployeeDep = "EmployeeDep4" };
            _pvtList.Add(mod1);
            _pvtList.Add(mod2);
            _pvtList.Add(mod3);
            _pvtList.Add(mod4);

            return _pvtList;
        }

    }
}

You can observe that Default Action method is creating a list of employees and send it back to view.


Step 3: Create the View representing the Grid in a very creative manner. While creating the different fields of the grid, ensure the Ids are unique and relative to one component of model.
Here i consider EmpId to be the Key Fields of reference for all the rows.
See how i created the grid taking care of above functionality.

@model IEnumerable<MVC3SimpleGrid.Models.EmployeeModel>
@{
    ViewBag.Title = "Index";
}
<div id="divResult">
<h2>Simple MVC Grid with Row Level Edit & Save</h2>
<table>
<tr>
<th>Employee ID</th>
<th>Employee Name</th>
<th>Employee Department</th>
<th>Action</th>
</tr>
@foreach (var m in Model)
{
    <tr>
        <td>
            @m.EmpId
        </td>
        <td>
            <div id="divEmpName-@m.EmpId.ToString()" class="visible">@m.EmployeeName</div>
            <input id="txtEmpName-@m.EmpId.ToString()" type="text" class="hide" value=@m.EmployeeName.ToString() />
        </td>
        <td>
            <div id="divEmpDep-@m.EmpId.ToString()"  class="visible">@m.EmployeeDep</div>
            <input id="txtEmpDep-@m.EmpId.ToString()" type="text"  class="hide" value=@m.EmployeeDep.ToString() />
        </td>
        <td>
            <button id="btnEdit-@m.EmpId.ToString()"  class="visible" onclick="ShowEdit(@m.EmpId); return false;">Edit</button>
            <button id="btnSave-@m.EmpId.ToString()"  class="hide" onclick="SaveEdit(@m.EmpId); return false;">Save</button>
        </td>
    </tr>
}
</table>
</div>

Immediately run the view in browser and see how the Ids are generated.

<div id="divResult">
<h2>Simple MVC Grid with Row Level Edit & Save</h2>
<table>
<tr>
<th>Employee ID</th>
<th>Employee Name</th>
<th>Employee Department</th>
<th>Action</th>
</tr>
    <tr>
        <td>
            1
        </td>
        <td>
            <div id="divEmpName-1" class="visible">Employee1</div>
            <input id="txtEmpName-1" type="text" class="hide" value=Employee1 />
        </td>
        <td>
            <div id="divEmpDep-1"  class="visible">EmployeeDep1</div>
            <input id="txtEmpDep-1" type="text"  class="hide" value=EmployeeDep1 />
        </td>
        <td>
            <button id="btnEdit-1"  class="visible" onclick="ShowEdit(1); return false;">Edit</button>
            <button id="btnSave-1"  class="hide" onclick="SaveEdit(1); return false;">Save</button>
        </td>
    </tr>
 . . . AND SO ON . . . 

Step 4: Here comes the JQuery part. We need to create the script to address the functions attached to both the buttons to do their respective tasks.

<script type="text/javascript">
    function ShowEdit(par) {
        $("#divEmpName-" + par).attr("class", "hide");
        $("#txtEmpName-" + par).attr("class", "visible");
        $("#divEmpDep-" + par).attr("class", "hide");
        $("#txtEmpDep-" + par).attr("class", "visible");
        $("#btnEdit-" + par).attr("class", "hide");
        $("#btnSave-" + par).attr("class", "visible");
    }

    function SaveEdit(par) {
        $("#divEmpName-" + par).attr("class", "visible");
        $("#txtEmpName-" + par).attr("class", "hide");
        $("#divEmpDep-" + par).attr("class", "visible");
        $("#txtEmpDep-" + par).attr("class", "hide");
        $("#btnEdit-" + par).attr("class", "visible");
        $("#btnSave-" + par).attr("class", "hide");

        var _empName = $("#txtEmpName-" + par).val();
        var _empDep = $("#txtEmpDep-" + par).val();
        var url = '@Url.Action("Index","Home")';
        $.post(url, { Id: par, empName: _empName, empDep: _empDep },
            function (data) {
               
                $("#divResult").html(data);
            });
    }    
</script>

Step 5: Everything is there except the Ajax version of Index Action method for supporting the SaveEdit(). Below is the implementation of it.

[HttpPost]
        public ActionResult Index(int Id,string empName,string empDep)
        {
            _lstEmployee[Id - 1].EmployeeName = empName;
            _lstEmployee[Id - 1].EmployeeDep = empDep;
            return View(_lstEmployee);
        }

We are all set with coding part. Now execute and see the results.
Results: