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.
Let's 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
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Please note that this application is purely to demonstrate how simple it is to create the GridView in MVC 3.
Code:
Html GridView in MVC3
Is it helpful for you? Kindly let me know your comments / Questions.