android - 将参数传递给BroadcastReceiver

我正在尝试将一个int值从我的Service传递给CallReceiver类，不幸的是，即使在设置了另一个值之后，CallReceiver.value也始终等于0。当我尝试将其作为参数传递给构造函数时，情况完全相同，因此从service调用的setter方法也是如此。真的没有办法传递任何数据吗？

服务:

SharedPreferences settings = getSharedPreferences("SETTINGS", 0);
    int value = settings.getInt("value1", 0); // here the correct value is present, not 0.
    CallReceiver mCallReceiver = new CallReceiver();
    CallReceiver.value = value;

接收者:

public class CallReceiver extends BroadcastReceiver  {

public int value;

public CallReceiver(int value)  {
    this.value = value;
}

public CallReceiver()   {

}


@Override
public void onReceive(Context context, Intent intent) {


            Log.v("value", String.valueOf(value)); // here "value" = 0.


        }

     }

最佳答案

您的CallReceiver mCallReceiver=new CallReceiver();实例不用于接收意图。相反，Android每次都会创建一个新实例。 0是未初始化整数变量的默认值。

为了确保发生这种情况，请为value字段分配一些默认值:

public class RReceiver extends BroadcastReceiver {
    public int value=5;
    //...
}

并且您的值将始终等于5。

至于将数据传递到BroadcastReceiver，请将其额外添加到您正在广播的Intent中:

//in your service
Intent broadcastedIntent=new Intent(this, CallReceiver.class);
broadcastedIntent.putExtra("VALUE", 100500);
sendBroadcast(broadcastedIntent);

然后，在您的CallReceiver中:

@Override
public void onReceive(Context context, Intent intent) {
    int value=intent.getIntExtra("VALUE", 0);
}