请使用五(5)个参数的正确重载方法。例子:

@using (@Ajax.BeginForm("Register", "Account", null,
    new AjaxOptions
    {
        HttpMethod = "POST",
        OnSuccess = "OnSuccess",
        OnFailure = "OnFailure",
        OnBegin = "OnBegin",
        OnComplete = "OnComplete"
    }, new { @class = "form-login" }))

这工作得很好

@Html.ActionLink("Informationen", "About", "Home", new { area = "" }, new { @class = "nav-link" })

新增{area = ""}。

也许其他人也有同样的问题，需要通过HTMLAttributes parm提供一个类值。 以下是我的解决方案:

@Html。ActionLink("About"， "About"， new {controller = "Home"， area = ""}， new {hidefocus = "hidefocus"， @class = "nav-item nav-link"})

只需删除“Home”(控制器的名称)，这样代码将是:

Html.ActionLink("About", "About", new { hidefocus = "hidefocus" })

我解决这个问题的方法是在匿名声明(new{})之前添加一个null到第四个参数，以便它使用以下方法重载:(linkText, actionName, controllerName, routeValues, htmlAttributes):

Html.ActionLink("About", "About", "Home", null, new { hidefocus = "hidefocus" })

属性名:

 @Html.ActionLink(linkText: "SomeText", actionName: "SomeAction", controllerName: "SomeControllerName", routeValues: new { parameterName = parameterValue}, htmlAttributes: null)