Question

我希望Logout链接执行位于HomeController的流程，而不是显示新的View。我该如何构建这个？

控制器方法：

public ActionResult LogoutProcess()
    {
        previousLoggedIn = WebSecurity.CurrentUserName;
        WebSecurity.Logout();
        return RedirectToAction("Logout", "Home");
    }

    public ActionResult Logout(HomeModels.LogoutModel model)
    {
        model.PreviouslyLoggedInUsername = previousLoggedIn;
        return View(model);
    }

查看：

<a href = "@Url.Action("LogoutProcess", "Home")">Logout</a>

Answer 1

您可以使用标准链接，定位该操作

@Url.Action("LogoutProcess", "Home")

“技巧”在重定向中，在LogoutProcess()操作结束时显示在其他视图中：

public ActionResult LogoutProcess()
{
    // TempData to transfer user name      
    TempData["previousLoggedIn"] = WebSecurity.CurrentUserName;
    WebSecurity.Logout();
    return RedirectToAction("Logout", "Home");
}

public ActionResult Logout(HomeModels.LogoutModel model)
{
    // fill model from TempData
    model.PreviouslyLoggedInUsername = TempData["previousLoggedIn"];
    return View(model);
}

CurrentUserName通过TempData

传递给其他操作

Answer 2

试试这个：

public ActionResult LogoutProcess()
{
    WebSecurity.Logout();
    //return null;
    Return RedirectToAction("Index");//or whatever page you want to display after logout.
}

Answer 3

您是否考虑过无内容的http状态代码结果？

return new HttpStatusCodeResult(HttpStatusCode.NoContent);

想要超链接到控制器方法而不是视图

3 个答案: