我也遇到过这个“问题”，但通过添加name属性找到了一个相当合理的解决方案。我不记得在其他语言中有这个问题。

http://www.w3.org/TR/html401/interact/forms.html#h-17.13.2

.．. 如果表单包含多个提交按钮，则只有激活的提交按钮是成功的。 .．.

这意味着以下代码值属性可以被更改、本地化、国际化，而不需要额外的代码检查强类型资源文件或常量。

<% Html.BeginForm("MyAction", "MyController", FormMethod.Post); %>
<input type="submit" name="send" value="Send" />
<input type="submit" name="cancel" value="Cancel" />
<input type="submit" name="draft" value="Save as draft" />
<% Html.EndForm(); %>`

在接收端，您只需要检查是否有任何已知的提交类型不是空

public ActionResult YourAction(YourModel model) {

    if(Request["send"] != null) {

        // we got a send

    }else if(Request["cancel"]) {

        // we got a cancel, but would you really want to post data for this?

    }else if(Request["draft"]) {

        // we got a draft

    }

}

你应该能够为按钮命名并赋予它们一个值;然后将此名称映射为操作的参数。或者，使用2个单独的动作链接或2个表单。

以下是最适合我的方法:

<input type="submit" value="Delete" name="onDelete" />
<input type="submit" value="Save" name="onSave" />


public ActionResult Practice(MyModel model, string onSave, string onDelete)
{
    if (onDelete != null)
    {
        // Delete the object
        ...
        return EmptyResult();
    }

    // Save the object
    ...
    return EmptyResult();
}

对于每个提交按钮，只需添加:

$('#btnSelector').click(function () {

    $('form').attr('action', "/Your/Action/);
    $('form').submit();

});

Modified version of HttpParamActionAttribute method but with a bug fix for not causing an error on expired/invalid session postbacks. To see if this is a problem with your current site, open the your form in a window and just before you go to click Save or Publish, open a duplicate window, and logout. Now go back to your first window and try to submit your form using either button. For me I got an error so this change solves that problem for me. I omit a bunch of stuff for the sake of brevity but you should get the idea. The key parts are the inclusion of ActionName on the attribute and making sure the name passed in is the name of the View that shows the form

属性类

[AttributeUsage(AttributeTargets.Method, AllowMultiple = false, Inherited = true)]
public class HttpParamActionAttribute : ActionNameSelectorAttribute
{
    private readonly string actionName;

    public HttpParamActionAttribute(string actionName)
    {
        this.actionName = actionName;
    }

    public override bool IsValidName(ControllerContext controllerContext, string actionName, MethodInfo methodInfo)
    {
        if (actionName.Equals(methodInfo.Name, StringComparison.InvariantCultureIgnoreCase))
            return true;

        if (!actionName.Equals(this.actionName, StringComparison.InvariantCultureIgnoreCase))
            return false;

        var request = controllerContext.RequestContext.HttpContext.Request;
        return request[methodInfo.Name] != null;
    }
}

控制器

[Authorize(Roles="CanAddContent")]
public ActionResult CreateContent(Guid contentOwnerId)
{
    var viewModel = new ContentViewModel
    {
        ContentOwnerId = contentOwnerId
        //populate rest of view model
    }
    return View("CreateContent", viewModel);
}

[Authorize(Roles="CanAddContent"), HttpPost, HttpParamAction("CreateContent"), ValidateAntiForgeryToken]
public ActionResult SaveDraft(ContentFormModel model)
{
    //Save as draft
    return RedirectToAction("CreateContent");
}

[Authorize(Roles="CanAddContent"), HttpPost, HttpParamAction("CreateContent"), ValidateAntiForgeryToken]
public ActionResult Publish(ContentFormModel model)
{
    //publish content
    return RedirectToAction("CreateContent");
}

View

@using (Ajax.BeginForm("CreateContent", "MyController", new { contentOwnerId = Model.ContentOwnerId }))
{
    @Html.AntiForgeryToken()
    @Html.HiddenFor(x => x.ContentOwnerId)

    <!-- Rest of your form controls -->
    <input name="SaveDraft" type="submit" value="SaveDraft" />
    <input name="Publish" type="submit" value="Publish" />
}

我尝试综合所有的解决方案，并创建了一个[ButtenHandler]属性，它可以很容易地处理表单上的多个按钮。

我已经在CodeProject中描述了它在ASP中的多个参数化(本地化)表单按钮。净MVC。

要处理此按钮的简单情况:

<button type="submit" name="AddDepartment">Add Department</button>

你会有如下的动作方法:

[ButtonHandler()]
public ActionResult AddDepartment(Company model)
{
    model.Departments.Add(new Department());
    return View(model);
}

注意按钮的名称与动作方法的名称是如何匹配的。本文还描述了如何让按钮具有值和按钮具有索引。