Action Method Parameters In ASP.NET MVC

Action Method Parameters are most important in MVC. If you want to handle post request in action methods; MVC framework provided types of Action Methods Parameters.

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.

1. Form Collection
2. Formal Parameters
3. 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.