JS 跳转后保持当前参数

跳转
                var data = str+"&jc=0"+ "&data=" + $("form").serialize().replace(/&/g, "(").replace(/=/g, ")");
                window.location.href = data;
接收
          var data1 = data.Replace("(", "&").Replace(")", "=");
            ViewBag.URL = "/Backxx/xxx/List/?id=5001" + "&" + data1; //调回到菜单

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转载自www.cnblogs.com/enych/p/11691168.html