从由BroadcastReceiver启动的服务更新ProgressBar [英] Update ProgressBar from a Service started by a BroadcastReceiver

问题描述

这是场景:我有一个BroadcastAction接收器，它监听ACTION_PHONE_STATE_CHANGED，并且当它更改时，它将启动服务.该服务通过CallLog进行查询，当查询增加时，我想在MainActivity的几个ProgressBar上显示它，但是我该怎么做呢?

This is the scenario: i have a BroadcastReceiver that listen to the ACTION_PHONE_STATE_CHANGED, and when it changes it starts a Service. This service make a query con the CallLog, and when something increases i want to show it on several ProgressBar on the MainActivity, but how can i do that?

我正在考虑创建一个Application类，该类将MainActivity和Service子类化，然后我可以拥有相应的引用，或者我错了吗?

I was thinking on creating an Application class thats subclasses the MainActivity and the Service, and then i could have the corresponding references, or am i wrong?

预先感谢

推荐答案

有一些可能性:

1. 您的服务可以发送广播( sendBroadcast )，您的活动仅需根据您在服务上实现的方式注册一个broacastReceiver.像这样的东西:

1. Your Service can send a brodcast(sendBroadcast) and your activity just need register a broacastReceiver according your implemented on your service. Something like that:

您的服务

 Intent intent=new Intent(getApplicationContext(),WebResults.class);
 intent.setAction("com.toxy.LOAD_URL");
 intent.putExtra("url",uri.toString());
 sendBroadcast(intent);

您的活动:

 private class Receiver extends BroadcastReceiver {

 @Override
 public void onReceive(Context arg0, Intent arg1) {

     String url = arg1.getExtras().getString("url");
     WebView webview =(WebView)findViewById(R.id.webView);
     webview.loadUrl(url);
 }

此链接可以帮助您: http://www.vogella.com/articles/AndroidBroadcastReceiver/article.html

第二种方法是在您的服务上注册一个侦听器，并在您的服务上通知该侦听器.

The second way is register a listener on your service and notify that listener on your service.

为您服务:

void onSomethingHappened()
{
     activityListener.notifyProgressBar();
}

void subscribe(YourInterface listener)
{
     actibityListener =  listener;
}

在您的活动中，开始服务后:

In you activity, after start you service:

service.subscribe(this); 

但是，要采用更好的解决方案，您应该考虑您的服务实现(类型).

However, to take the better solution you should take into account your service implementation(type).

关于 AndroidService 的链接可能会有用.

This link about AndroidService can be useful.

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