您好,我正在开发一个应用程序,当耳机从手机上取下时,该应用程序会生成一个事件。
我创建了一个广播接收器,其接收方法为
public void onReceive(Context context, Intent intent) {
// TODO Auto-generated method stub
String action = intent.getAction();
Log.i("Broadcast Receiver", "Hello");
if( (action.compareTo(Intent.ACTION_HEADSET_PLUG)) == 0) //if the action match a headset one
{
int headSetState = intent.getIntExtra("state", 0); //get the headset state property
int hasMicrophone = intent.getIntExtra("microphone", 0);//get the headset microphone property
if( (headSetState == 0) && (hasMicrophone == 0)) //headset was unplugged & has no microphone
{
//do whatever
}
}
}
调用该方法如下
IntentFilter receiverFilter = new IntentFilter(Intent.ACTION_HEADSET_PLUG);
HeadSetBroadCastReceiver receiver = new HeadSetBroadCastReceiver();
registerReceiver( receiver, receiverFilter );
我也已在清单中将其注册为
<receiver android:name=".HeadsetBroadCastReceiver">
<intent-filter>
<action android:name="android.intent.action.ACTION_HEADSET_PLUG"/>
</intent-filter>
</receiver>
和许可
但这不起作用任何人都可以指导我完成这个吗?
这是棘手的一点,但您可以使用广播作为以下内容与我配合良好
在你的Activity
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
myReceiver = new HeadSetReceiver();
}
and in onResume()方法注册您的广播
public void onResume() {
IntentFilter filter = new IntentFilter(Intent.ACTION_HEADSET_PLUG);
registerReceiver(myReceiver, filter);
super.onResume();
}
然后在您的活动中声明您的广播
private class HeadSetReceiver extends BroadcastReceiver {
@Override public void onReceive(Context context, Intent intent) {
if (intent.getAction().equals(Intent.ACTION_HEADSET_PLUG)) {
int state = intent.getIntExtra("state", -1);
switch (state) {
case 0:
Log.d(TAG, "Headset unplugged");
break;
case 1:
Log.d(TAG, "Headset plugged");
break;
}
}
}
}
希望有帮助,
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