这期内容当中小编将会给大家带来有关如何在Asp.net中利用mvc 实现一个超时弹窗后跳转功能,文章内容丰富且以专业的角度为大家分析和叙述,阅读完这篇文章希望大家可以有所收获。
public class PowerFilter : AuthorizeAttribute { public override void OnAuthorization(AuthorizationContext filterContext) { var cookie = HttpContext.Current.Request.Cookies["loginInfo"]; if(null == cookie) { filterContext.Result = new RedirectResult("/admin/login/index"); } else { cookie.Expires = DateTime.Now.AddMinutes(30); HttpContext.Current.Response.Cookies.Remove("loginInfo"); HttpContext.Current.Response.Cookies.Add(cookie); } } }
但是页面直接跳转了,也没有一个提示,显得不是很友好,可以这样
public class PowerFilter : AuthorizeAttribute { public override void OnAuthorization(AuthorizationContext filterContext) { var cookie = HttpContext.Current.Request.Cookies["loginInfo"]; if(null == cookie) { filterContext.Result = new ContentResult() { Content = string .Format("<script>alert('登录超时,请重新登录');location.href='{0}'</script>","/admin/login/index") }; } else { cookie.Expires = DateTime.Now.AddMinutes(30); HttpContext.Current.Response.Cookies.Remove("loginInfo"); HttpContext.Current.Response.Cookies.Add(cookie); } } } }
但是,假如是ajax请求呢?
public class PowerFilter : AuthorizeAttribute { public override void OnAuthorization(AuthorizationContext filterContext) { var cookie = HttpContext.Current.Request.Cookies["loginInfo"]; if(null == cookie) { if(!filterContext.HttpContext.Request.IsAjaxRequest()) { filterContext.Result = new ContentResult() { Content = string .Format("<script>alert('登录超时,请重新登录');location.href='{0}'</script>","/admin/login/index") }; } else { filterContext.Result = new JsonResult() { Data = new { logoff = true,logurl = "/admin/login/index" }, ContentType = null, ContentEncoding = null, JsonRequestBehavior = JsonRequestBehavior.AllowGet }; } } else { cookie.Expires = DateTime.Now.AddMinutes(30); HttpContext.Current.Response.Cookies.Remove("loginInfo"); HttpContext.Current.Response.Cookies.Add(cookie); } } }
上述就是小编为大家分享的如何在Asp.net中利用mvc 实现一个超时弹窗后跳转功能了,如果刚好有类似的疑惑,不妨参照上述分析进行理解。如果想知道更多相关知识,欢迎关注亿速云行业资讯频道。
免责声明:本站发布的内容(图片、视频和文字)以原创、转载和分享为主,文章观点不代表本网站立场,如果涉及侵权请联系站长邮箱:is@yisu.com进行举报,并提供相关证据,一经查实,将立刻删除涉嫌侵权内容。