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.
  1. using System.Web.Mvc;
  2. namespace MVCActionMethodParameters.Controllers
  3. {
  4. public class HomeController : Controller
  5. {
  6. // GET: Home
  7. [HttpGet]
  8. public ActionResult Index()
  9. {
  10. return View();
  11. }
  12. [HttpPost]
  13. public ActionResult Index(FormCollection frmobj) //FormCollection
  14. {
  15. string username = frmobj["userid"];
  16. string password = frmobj["pwd"];
  17. if (username == "Admin" && password == "123456")
  18. {
  19. Response.Write("<h2> Success </h2> Valid User...");
  20. }
  21. else
  22. Response.Write(" <h2> Failed </h2> Invalid User...");
  23. return View();
  24. }
  25. }
Step 4 - Right click on FormCollection Method and add view. Afterwards, create login fieldsm as shown below.
  1. @{
  2. ViewBag.Title = "Index";
  3. }
  4. <h1>Form Collection</h1>
  5. <h2>Login </h2>
  6. @using (Html.BeginForm())
  7. {
  8. <label>User Name</label>
  9. <input type="text" class="form-control" id="userid" name="userid" />
  10. <label>Password</label>
  11. <input type="text" class="form-control" id="pwd" name="pwd" /> <br />
  12. <input type="submit" class="btn btn-success" value="Submit" />
  13. }
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.
  1. using System.Web.Mvc;
  2. namespace MVCActionMethodParameters.Controllers
  3. {
  4. public class LoginController : Controller
  5. {
  6. // GET: Login
  7. [HttpGet]
  8. public ActionResult Login()
  9. {
  10. return View();
  11. }
  12. [HttpPost]
  13. public ActionResult Login(string userid, string pwd) //Formal Parameters
  14. {
  15. string username = userid;
  16. string password = pwd;
  17. if (username == "Admin" && password == "123456")
  18. {
  19. Response.Write("<h2> Success </h2> Valid User...");
  20. }
  21. else
  22. Response.Write(" <h2> Failed </h2> Invalid User...");
  23. return View();
  24. }
  25. }
  26. }
Step 6 - Right click on Formal paratmeter action method and add same login fields in View.
  1. @{
  2. ViewBag.Title = "Login";
  3. }
  4. <h1>Formal Parameters</h1>
  5. <hr />
  6. <h2>Login</h2>
  7. @using (Html.BeginForm())
  8. {
  9. <label>User Name</label>
  10. <input type="text" class="form-control" id="userid" name="userid" />
  11. <label>Password</label>
  12. <input type="text" class="form-control" id="pwd" name="pwd" /> <br />
  13. <input type="submit" class="btn btn-success" value="Submit" />
  14. }
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.
  1. namespace MVCActionMethodParameters.Models
  2. {
  3. public class Login
  4. {
  5. public string userid { get; set; }
  6. public string pwd { get; set; }
  7. }
  8. }
Step 9 - Again add another controller for Model Class Object and write the code given below.

  1. using System.Web.Mvc;
  2. using MVCActionMethodParameters.Models;
  3. namespace MVCActionMethodParameters.Controllers
  4. {
  5. public class LoginModelController : Controller
  6. {
  7. // GET: LoginModel
  8. public ActionResult LoginModel()
  9. {
  10. return View();
  11. }
  12. [HttpPost]
  13. public ActionResult LoginModel(Login obj) // Model class object
  14. {
  15. string username = obj.userid;
  16. string password = obj.pwd;
  17. if (username == "Admin" && password == "123456")
  18. {
  19. Response.Write("<h2> Success </h2> Valid User...");
  20. }
  21. else
  22. Response.Write(" <h2> Failed </h2> Invalid User...");
  23. return View();
  24. }
  25. }
  26. }
Step 10 - Right click on Model class object method and add the view.
  1. <h1>Formal Parameters</h1>
  2. <hr />
  3. <h2>Login</h2>
  5. @using (Html.BeginForm())
  6. {
  7. <label>User Name</label>
  8. <input type="text" class="form-control" id="userid" name="userid" />
  9. <label>Password</label>
  10. <input type="text" class="form-control" id="pwd" name="pwd" /> <br />
  11. <input type="submit" class="btn btn-success" value="Submit" />
  12. }
Step 11 - Run the Application. Enter a valid userid and password. Now, you will see the output.