从动作脚本类推送视图

Question

我有一个远程对象类，用于处理与远程数据服务的所有交互。每次进行调用时，数据服务都会检查用户是否具有有效的会话。如果不这样做，则不会运行请求的方法并返回失败。我可以在故障处理程序中捕获此故障。如果发生这种情况，我想做的就是将登录屏幕推送给用户。

我已经尝试了以下方法

            var navigator:ViewNavigator;
            navigator.activeView.navigator.pushView(views.LoginScreen);

，但这不起作用，并且失败，无法访问空对象引用的属性或方法。这是有道理的。所以我的问题是如何获取当前正在运行的视图导航器对象的引用并推送视图？

谢谢

这里要求的是完整的远程对象类

包远程处理 { 导入事件.RemoteExceptionEvent;

import flash.events.*;

import mx.managers.CursorManager;
import mx.messaging.ChannelSet;
import mx.messaging.channels.AMFChannel;
import mx.rpc.events.FaultEvent;
import mx.rpc.remoting.RemoteObject;

import spark.components.ViewNavigator;

import views.FirstTime.ValidateUser;

/**
 * Super class for all remote services that contains some generic methods.
 */
public class RemoteService extends EventDispatcher

{
    private static var REMOTE_EXCEPTION:String = "Remote exception";
    private static var NO_MESSAGE:String = "10001";

    protected var remoteObject:RemoteObject;

    private var amfChannelSet:ChannelSet;

    /**
     * Constructor accepting an id and destination for the actual RemoteObject to create. An event listener
     * is added for exceptions.
     * @param id String representing the id of the new RemoteObject to create
     * @param destination String representing the destination of the RemoteObject to create
     * @param amfChannelId String representing the Channel of the RemoteObject to create
     * @param amfChannelEndpoint String representing the Endpoint URI of the RemoteObject to create
     */
    public function RemoteService( serviceId:String
                                   , serviceDestination:String
                                     , amfChannelId:String
                                       , amfChannelEndpoint:String
    ) 
    {
        // Create a runtime Channelset for given Channel ID and Endpoinr URI
        var amfChannel:AMFChannel = new AMFChannel(amfChannelId, amfChannelEndpoint);
        amfChannelSet = new ChannelSet();
        amfChannelSet.addChannel(amfChannel);

        // Create the remoteObject instance
        this.remoteObject = new RemoteObject(serviceId);
        this.remoteObject.channelSet = amfChannelSet;
        this.remoteObject.destination = serviceDestination;
        this.remoteObject.addEventListener(FaultEvent.FAULT,onRemoteException);
        this.remoteObject.setCredentials('test','test');

    }

    /**
     * generic fault event handler for all remote object actions. based on the received message/code an action
     * is taken, mostly throwing a new event.
     * @param event FaultEvent received for handling
     */
    public function onRemoteException(event:FaultEvent):void {
        trace('code : ' + event.fault.faultCode +
            ', message : ' + event.fault.faultString +
            ',detail : ' + event.fault.faultDetail);

            trace('fodun: ' + event.fault.faultDetail.indexOf("Authentication"));

        if (event.fault.faultDetail.indexOf("Authentication") > 0)
        {

            var navigator:ViewNavigator;

             navigator.activeView.navigator.pushView(views.LoginScreen);



        }
        else  if (event.fault.faultString == REMOTE_EXCEPTION) {
            EventDispatcher(
                new RemoteExceptionEvent(RemoteExceptionEvent.REMOTE_EXCEPTION,
                    "unknown problem occurred during a remote call : " + event.fault.message));
        } else if (event.fault.faultCode == NO_MESSAGE) { 
            EventDispatcher(
                new RemoteExceptionEvent(RemoteExceptionEvent.REMOTE_EXCEPTION,
                    event.fault.faultString));
        } else {

            EventDispatcher((
                new RemoteExceptionEvent(RemoteExceptionEvent.REMOTE_EXCEPTION,
                    "unknown runtime problem occurred during a remote call : " + event.fault.message)));
        }
    }
}
}

Answer 1

从 RemoteService 类中访问视图或任何可视组件并不是处理 MVC 应用程序的好方法。您想要做的是在 RemoteService 类中调度一个事件，该事件基本上表示“身份验证失败”。然后，在您看来，您将侦听此事件并将其解释为显示登录屏幕的需要。看 ？