Create strong type view in MVC application

Sourav.Kayal
Posted by in ASP.NET MVC category on for Beginner level | Points: 250 | Views : 14877 red flag

How to create strong type view in MVC appllication

Create strong type view in MVC application

In this quick article we will learn how to create strong type view in MVC application. There is little advantage in strong type view. We can send data in form of object. Let’s implement strong type view with small example.

Create simple model

Open one empty MVC application and add below model into it.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
 
namespace MVCTest.Models
{
    public class Person
    {
        public String Name{ get; set; }
        public String Surname { get; set; }
        public String Age { get; set; }
    }
}
 

Our model is very simple, only we have implemented Person classes which contain three properties.

Create Person Controller

Let’s create one controller to handle model data. As we have defined our model class name as Person. We will define controller name as PersonController. Have a look on below code.

using System;
using System.Collections.Generic;
using System.Linq;
using System.Web;
using System.Web.Mvc;
using MVCTest.Models;
 
namespace MVCTest.Controllers
{
    public class PersonController : Controller
    {
     
        public ActionResult Index()
        {
            return View();
        }
        public ActionResult GetData(Person objP)
        {
            String Name = objP.Name;
            String Surname = objP.Surname;
            Int32 Age = Convert.ToInt32(objP.Age);
            return Content("Name:-" + Name + " Surname:-" + objP.Surname + " Age:-" + objP.Age);
        }
 
    }
}

The controller contains tow actions. Index action will show HTML form, and GetData() function will call when we sill submit form.

Create strong type view

Now we will create strong type view. The advantage of strong type view is we can send and receive data in the form of object. To create strong type view, just right click on Index action, you will see below form.

 

Now,

 add below code in view

@model MVCTest.Models.Person
 
@{
    Layout = null;
}
 
<!DOCTYPE html>
 
<html>
<head>
    <title>Index</title>
</head>
<body>
    <div>
        @using (Html.BeginForm("Person", "GetData", FormMethod.Post))
        {
         @Html.LabelFor(m=>m.Name)
         @Html.TextBoxFor(m=> m.Name)
 
         @Html.LabelFor(m=>m.Surname)
         @Html.TextBoxFor(m => m.Surname);
                                                            @Html.LabelFor(m => m.Age)
         @Html.TextBoxFor(m => m.Age);
 
         <input type="submit" name="Submit" value="submit" />
        }
 
    </div>
</body>
</html>

Have a look on first line of View. We are pointing Person class as model. And as it is strong type view, we can use lamda expression to point each property of model class.


Output:-



Conclusion:-
In this article we have learned how to create strong type view in MVC application.
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About the Author

Sourav.Kayal
Full Name: Sourav Kayal
Member Level: Silver
Member Status: Member,MVP
Member Since: 6/20/2013 2:09:01 AM
Country: India
Read my blog here http://ctrlcvprogrammer.blogspot.in/
http://www.dotnetfunda.com
I am .NET developer working for HelixDNA Technologies,Bangalore in healthcare domain. Like to learn new technology and programming language. Currently working in ASP.NET ,C# and other microsoft technologies.

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