Action Method Parameters In ASP.NET MVC

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Action Method Parameters In ASP.NET MVC

Action Method Parameters

We can organize the action methods for GET and POST requests separately.
We can easily create seperate action methods for each request types.

Request Type Attribute

[HttpGet]
[HttpPost]

Action method parameters

MVC framework provides 3 different action methods for Post Request, which are given below.

Form Collection
Formal Parameters
Model Class Object

i) Form Collection

It is a pre-defined class in syste.web.mvc namespace.
It stores the items in key value pairs format only.

key - name of the input fields.
value - input content field (string type).

Step 1 - Create a new empty MVC Application.

Step 2 - Add the controller, as shown below.

Step 3 - First is controller example of FormCollection.

using System.Web.Mvc;
namespace MVCActionMethodParameters.Controllers
{
public class HomeController : Controller
{
// GET: Home
[HttpGet]
public ActionResult Index()
{
return View();
}
[HttpPost]
public ActionResult Index(FormCollection frmobj) //FormCollection
{
string username = frmobj["userid"];
string password = frmobj["pwd"];
if (username == "Admin" && password == "123456")
{
Response.Write("<h2> Success </h2> Valid User...");
}
else
Response.Write(" <h2> Failed </h2> Invalid User...");
return View();
}
}

Step 4 - Right click on FormCollection Method and add view. Afterwards, create login fieldsm as shown below.

@{
ViewBag.Title = "Index";
}
<h1>Form Collection</h1>
<h2>Login </h2>
@using (Html.BeginForm())
{
<label>User Name</label>
<input type="text" class="form-control" id="userid" name="userid" />
<label>Password</label>
<input type="text" class="form-control" id="pwd" name="pwd" /> <br />
<input type="submit" class="btn btn-success" value="Submit" />
}

ii) Formal Parameters

Using this option, we can get the values easily in the form of pre-defined data types example like int, string etc.
We can easily handle type casting.
It will read based on the form input fields name directly.

Note - The name of the parameters should be the same as the input field name. Afterwards, it will possible to achieve the formal parameters.

Step 5 - Again add another controller, give any name of the controller and paste the code given below into Controller.

using System.Web.Mvc;
namespace MVCActionMethodParameters.Controllers
{
public class LoginController : Controller
{
// GET: Login
[HttpGet]
public ActionResult Login()
{
return View();
}
[HttpPost]
public ActionResult Login(string userid, string pwd) //Formal Parameters
{
string username = userid;
string password = pwd;
if (username == "Admin" && password == "123456")
{
Response.Write("<h2> Success </h2> Valid User...");
}
else
Response.Write(" <h2> Failed </h2> Invalid User...");
return View();
}
}
}

Step 6 - Right click on Formal paratmeter action method and add same login fields in View.

@{
ViewBag.Title = "Login";
}
<h1>Formal Parameters</h1>
<hr />
<h2>Login</h2>
@using (Html.BeginForm())
{
<label>User Name</label>
<input type="text" class="form-control" id="userid" name="userid" />
<label>Password</label>
<input type="text" class="form-control" id="pwd" name="pwd" /> <br />
<input type="submit" class="btn btn-success" value="Submit" />
}

iii ) Model class object

It is the same as business entities.
We have created a seperate class for the objects in model folder.
In MVC model, logic is prepared by using the classes, which are based on the database tables.
It requires model class objects.

Note - The name of the parameters should be same as the input field name, then only it will possible to achieve formal parameters.

Step 7 - Right click on Model folder and add the class.

Step 8 - Create a class with the objects.

namespace MVCActionMethodParameters.Models
{
public class Login
{
public string userid { get; set; }
public string pwd { get; set; }
}
}

Step 9 - Again add another controller for Model Class Object and write the code given below.

using System.Web.Mvc;
using MVCActionMethodParameters.Models;
namespace MVCActionMethodParameters.Controllers
{
public class LoginModelController : Controller
{
// GET: LoginModel
public ActionResult LoginModel()
{
return View();
}
[HttpPost]
public ActionResult LoginModel(Login obj) // Model class object
{
string username = obj.userid;
string password = obj.pwd;
if (username == "Admin" && password == "123456")
{
Response.Write("<h2> Success </h2> Valid User...");
}
else
Response.Write(" <h2> Failed </h2> Invalid User...");
return View();
}
}
}

Step 10 - Right click on Model class object method and add the view.

<h1>Formal Parameters</h1>
<hr />
<h2>Login</h2>
@using (Html.BeginForm())
{
<label>User Name</label>
<input type="text" class="form-control" id="userid" name="userid" />
<label>Password</label>
<input type="text" class="form-control" id="pwd" name="pwd" /> <br />
<input type="submit" class="btn btn-success" value="Submit" />
}

Step 11 - Run the Application. Enter a valid userid and password. Now, you will see the output.