MVC - How to simply do something at button click? - asp.net-mvc

Question

I am new to ASP.NET MVC. I was used to program using just ASP.NET. I want to do something when the user clicks a button. I am not understanding what do I do at Controller.
I have this View:
#{
ViewBag.Title = "Index";
Layout = "~/Views/Shared/_Layout.cshtml";
}
#Html.Action("RegisterHour", "Register")
and the Controller:
public ActionResult RegisterHour()
{
//TODO: do anything
return View("Index");
}
When I click at the button, I would like to stay in the same page (it can reload). I simply want to do something like go to the database and create a new entity, and then show a messagebox.
This code causes an stackoverflow. What am I missing? What do I have to change at Controller?
Thanks.

Answer 1

The line
#Html.Action("RegisterHour", "Register")
actually makes a request to the server in order to render the result of the "RegisterHour" action. So in order to render the result of the action the first time, you need to make a request to the same action. This causes an endless loop, hence the stack overflow.
You are thinking in events, rather than thinking about HTTP and the web.
ASP.NET MVC embraces the HTTP protocol and you have to know what happens when a request is made and how HTML is rendered.
If you want to implement the scenario you are describing, you have to put a form on the page. The button can submit that form by making a POST request to some other action, and then the action can render a view showing the result. But for simply showing a message box, I don't think it is a good idea.
This is how desktop apps work, not web apps. You are trying to fit a square peg through a round hole.

Answer 2

there are many ways to get back to the controller. without post back look into ajax calls. the simplest way is the post back. put your view in a form tag either html or html.beginform
#using (Html.BeginForm()){
<input type="submit" value="submit"/>
}
as #Chuck mentioned on your controller then have a post method with the same name as the get you show
[HttpPost]
public ActionResult RegisterHour()
{
//TODO: do anything
return View(model);
}
the #Html.Action that you have will return a url so that is put inside another element. something like
<a src="#Html.Action("Action", "Controller")">Click Here</a>

Answer 3

if you want to stay in the same page instead of
return View("index");
use
return View();
Edit:
If you want a complete code of a do something stuff here you are:
Model
public ActionResult MyModel()
{
[Required]
public int propriety1 { get; set; }
}
Controller
public ActionResult DoSomething()
{
var model = new MyModel();
return View(model);
}
[HttpPost]
public ActionResult DoSomething(MyModel model)
{
if(ModelState.isValid){
//DO something
}else{
return View(model);
}
}
View
#model Models.MyModel
#using (Html.BeginForm())
{
#Html.LaberlFor(m=>m.propriety1) #Html.TextBoxFor(m=>m.propriety1)
<input type="submit" value="Click me" />
}

Answer 4

Create an endpoint and have the form submit to it.
UI Code
<form action="/Registration/RegisterHour" >
<p>
<label>Username</label>
<input type="text" name="username" />
</p>
<p>
<label>First Name</label>
<input type="text" name="firstname" />
</p>
<p>
<label>Last Name</label>
<input type="text" name="lastname" />
</p>
<p>
<label>Password</label>
<input type="text" name="password" />
</p>
<p>
<label>Confirm</label>
<input type="text" name="confirm" />
</p>
<input type="submit" value="Save" />
</form>
Model
public class Registration
{
public string Username {get; set;}
public string FirstName {get; set;}
public string LastName {get; set;}
public string Password {get; set;}
public string Confirm {get; set;}
}
Typically you'll have two of the same endpoints, one is for the get, the other is for the post.
public class RegistrationController : Controller
{
//Get
[HttpGet]
public ActionResult RegisterHour()
{
//TODO: do anything
return View("Index");
}
//Post
[HttpPost]
public ActionResult RegisterHour(Registration newUser)
{
if(Model.IsValid)
{
//Save user to the database
userRepository.AddUser(newUser);
//load success screen.
return RedirectAction("SuccessfulRegistration");
}
//If Model is invalid handle error on the client.
return View("Index");
}
}