returnUrl总是null？

Question

好吧，就像标题所说的那样，我的登录操作中的returnUrl参数始终为null，我遵循默认的Internet应用程序示例，但我无法使其工作（我对MVC相对较新）

视图

@using MundialDeFutbol.Models
@using System.Web.Optimization
@model LoginModel

@{
    ViewBag.Title = "Login";
}

<h3 class="text-info">Login</h3>
<div class="separadorHorizontal"></div>

@using (Html.BeginForm("Login", "Usuarios", new { ReturnUrl = ViewBag.ReturnUrl }, FormMethod.Post,
    new { @class = "form-horizontal", @style = "margin-top: 20px;" }))
{
    @Html.AntiForgeryToken()

    <div id="loginErrors" class="form-group">
        <div class="col-sm-offset-2 col-sm-10 ">
            <strong>
                @Html.ValidationSummary(true, null, new { @class = "text-danger", @id = "margin-top: 10px;" })
            </strong>
        </div>
    </div>
    <div class="form-group">
        <label for="Nick" class="col-sm-2 control-label">@Html.LabelFor(model => model.Nick)</label>
        <div class="col-sm-10">
            @Html.TextBoxFor(model => model.Nick, new { @class = "form-control", @placeholder = "Nombre de usuario" })
            <strong>
                @Html.ValidationMessageFor(model => model.Nick, null, new { @class = "text-danger", @style = "display: block; margin-top: 10px;" })
            </strong>
        </div>
    </div>
    <div class="form-group">
        <label for="Contraseña" class="col-sm-2 control-label">@Html.LabelFor(model => model.Contraseña)</label>
        <div class="col-sm-10">
            @Html.TextBoxFor(model => model.Contraseña, new { @class = "form-control", @placeholder = "Escribe contraseña", @type = "password" })
            <strong>
                @Html.ValidationMessageFor(model => model.Contraseña, null, new { @class = "text-danger", @style = "display: block; margin-top: 10px;" })
            </strong>
        </div>
    </div>
    <div class="form-group">
        <div class="col-sm-offset-2 col-sm-10">
            <button type="submit" class="btn btn-default">Entrar</button>
        </div>
    </div>
}

@section scripts {
    @Scripts.Render("~/bundles/jqueryval")
}

控制器

    [AllowAnonymous]
    public ActionResult Login(string returnUrl)
    {
        ViewBag.ReturnUrl = returnUrl;
        return View();
    }

    [HttpPost]
    [AllowAnonymous] // --> Preguntar
    [ValidateAntiForgeryToken] // --> Preguntar
    public ActionResult Login(LoginModel usuario, string returnUrl)
    {
        if (ModelState.IsValid)
        {
            if (IsValid(usuario.Nick,usuario.Contraseña))
            {
                FormsAuthentication.SetAuthCookie(usuario.Nick, false);
                return Redirect(returnUrl);
            }
            else
            {
                ModelState.AddModelError("", "Usuario o contraseña incorrectos");
            }
        }
        else
        {
            var errors = ModelState.Keys.SelectMany(key => ModelState[key].Errors);
        }

        return View();
    }

最后，我想问一下“[AllowAnonymous]”和“[ValidateAntiForgeryToken]”装饰器是什么。

提前致谢！

Answer 1

我猜你正在设置两件事但是用作一个：在你设置的.cshtml页面中：

@using (Html.BeginForm("Login", "Usuarios", new { ReturnUrl = ViewBag.ReturnUrl }, FormMethod.Post,
new { @class = "form-horizontal", @style = "margin-top: 20px;" }))

但是在你发布登录操作时，你给它的名字叫“returnUrl”，如下所示：

public ActionResult Login(LoginModel usuario, string returnUrl)

请检查并重试，参数区分大小写。

Answer 2

在参数中你有：

new { ReturnUrl = ViewBag.ReturnUrl }
      ^
      |
      |

但在行动参数是：

public ActionResult Login(LoginModel usuario, string returnUrl)
                                                     ^
                                                     |
                                                     |

Answer 3

如果您使用自定义身份验证，则需要添加方法OnAuthorization：ReturnUrl = filterContext.HttpContext.Request.RawUrl

示例：

 public override void OnAuthorization(AuthorizationContext filterContext)
 {

    //Some logic here...
    filterContext.Result = new RedirectToRouteResult(
                           new RouteValueDictionary(
                           new { controller = "Account",
                                             action = "Login", 
                                             ReturnUrl = filterContext.HttpContext.Request.RawUrl }));            
 }