How to link HTML5 form action to Controller Action

问题:

I have a basic form for which I want to handle buttons inside the form by calling the ActionResult method in the View's associated Controller class. Here is the following HTML5 code for the form:

<h2>Welcome</h2>

<div>

    <h3>Login</h3>

    <form method="post" action= <!-- what goes here --> >
        Username: <input type="text" name="username" /> <br />
        Password: <input type="text" name="password" /> <br />
        <input type="submit" value="Login">
        <input type="submit" value="Create Account"/>
    </form>

</div>

<!-- more code ... -->

The corresponding Controller code is the following:

[HttpPost]
public ActionResult MyAction(string input, FormCollection collection)
{
    switch (input)
    {
        case "Login":
            // do some stuff...
            break;
        case "Create Account"
            // do some other stuff...
            break;
    }

    return View();
}

回答1:

you make the use of the HTML Helper and have

    @using(Html.BeginForm())
    {
        Username: <input type="text" name="username" /> <br />
        Password: <input type="text" name="password" /> <br />
        <input type="submit" value="Login">
        <input type="submit" value="Create Account"/>
    }

or use the Url helper

<form method="post" action="@Url.Action("MyAction", "MyController")" >

Html.BeginForm has several (13) overrides where you can specify more information, for example, a normal use when uploading files is using:

@using(Html.BeginForm("myaction", "mycontroller", FormMethod.Post, new {enctype = "multipart/form-data"}))
{
    < ... >
}

---

If you don't specify any arguments, the Html.BeginForm() will create a POST form that points to your current controller and current action. As an example, let's say you have a controller called Posts and an action called Delete

public ActionResult Delete(int id)
{
   var model = db.GetPostById(id);
   return View(model);
}

[HttpPost]
public ActionResult Delete(int id)
{
    var model = db.GetPostById(id);
    if(model != null) 
        db.DeletePost(id);

    return RedirectToView("Index");
}

and your html page would be something like:

<h2>Are you sure you want to delete?</h2>
<p>The Post named <strong>@Model.Title</strong> will be deleted.</p>

@using(Html.BeginForm())
{
    <input type="submit" class="btn btn-danger" value="Delete Post"/>
    <text>or</text>
    @Url.ActionLink("go to list", "Index")
}

回答2:

Here I'm basically wrapping a button in a link. The advantage is that you can post to different action methods in the same form.

<a href="Controller/ActionMethod">
    <input type="button" value="Click Me" />
</a>

Adding parameters:

<a href="Controller/ActionMethod?userName=ted">
    <input type="button" value="Click Me" />
</a>

Adding parameters from a non-enumerated Model:

<a href="Controller/ActionMethod?userName=@Model.UserName">
    <input type="button" value="Click Me" />
</a>

You can do the same for an enumerated Model too. You would just have to reference a single entity first. Happy Coding!