如何将 HTML5 表单操作链接到 ASP.NET MVC 4 中的控制器 ActionResult 方法 [英] How to link HTML5 form action to Controller ActionResult method in ASP.NET MVC 4

问题描述

我有一个基本表单，我想通过调用视图关联的 Controller 类中的 ActionResult 方法来处理表单内的按钮.这是表单的以下 HTML5 代码:

欢迎光临
<div><h3>登录</h3><form method="post" action= <!-- 这里是什么-->>用户名:<input type="text" name="username"/><br/>密码:<input type="text" name="password"/><br/><input type="submit" value="登录"><input type="submit" value="创建账户"/></表单>

<!-- 更多代码...-->

对应的Controller代码如下:

[HttpPost]公共 ActionResult MyAction(字符串输入，FormCollection 集合){开关(输入){案例登录"://做一些事情...休息;案例创建帐户"//做一些其他的事情...休息;}返回视图()；}

解决方案

您使用 HTML Helper 并具有

 @using(Html.BeginForm()){用户名:<input type="text" name="username"/><br/>密码:<input type="text" name="password"/><br/><input type="submit" value="登录"><input type="submit" value="创建账户"/>}

或使用 Url 助手

Html.BeginForm 有几个 (13) 覆盖，您可以在其中指定更多信息，例如，上传文件时的正常使用正在使用:

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

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如果您不指定任何参数，Html.BeginForm() 将创建一个 POST 表单，该表单指向您的当前控制器和当前操作.例如，假设您有一个名为 Posts 的控制器和一个名为 Delete

的操作

public ActionResult Delete(int id){var 模型 = db.GetPostById(id);返回视图(模型)；}[HttpPost]公共操作结果删除(int id){var 模型 = db.GetPostById(id);如果(模型！= null)db.DeletePost(id);return RedirectToView("索引");}

而您的 html 页面将类似于:

您确定要删除吗?</h2><p>名为<strong>@Model.Title</strong>的帖子</p>@using(Html.BeginForm()){<input type="submit" class="btn btn-danger" value="删除帖子"/><text>或</text>@Url.ActionLink("转到列表", "索引")}

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();
}

解决方案

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")
}

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