如何以编程方式设置广播接收器属性？

Question

我正在我的应用程序中广播意图并使用广播接收器接收它。我可以处理广播和接收。没问题。但是，我想完全以编程方式注册接收器，而不是在清单文件中进行注册。请注意，在清单文件中，接收者有两个属性

android:enabled="true"

和

android:exported="false"

。我需要知道，当我以编程方式注册接收器时，如何具体设置这两个属性？

我的AndroidManifest.xml文件：

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
    package="com.example.mybroadcastapplication">

    <application
        android:allowBackup="true"
        android:icon="@mipmap/ic_launcher"
        android:label="@string/app_name"
        android:roundIcon="@mipmap/ic_launcher_round"
        android:supportsRtl="true"
        android:theme="@style/Theme.MyBroadcastApplication">
        <activity
            android:name=".MainActivity"
            android:exported="true">
            <intent-filter>
                <action android:name="android.intent.action.MAIN" />

                <category android:name="android.intent.category.LAUNCHER" />
            </intent-filter>
        </activity>
        <receiver
            android:name=".MyBroadcastReceiver"
            android:enabled="true"
            android:exported="false">
        </receiver>
    </application>

</manifest>

我的MainActivity.java文件：

public class MainActivity extends AppCompatActivity implements View.OnClickListener {
    MyBroadcastReceiver myReceiver;
    IntentFilter intentFilter;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);
        myReceiver = new MyBroadcastReceiver();
        intentFilter = new IntentFilter();
        intentFilter.addAction("com.example.mybroadcastapplication.EXPLICIT_INTENT");
        findViewById(R.id.button1).setOnClickListener(this);
    }

    public void broadcastIntent() {
        Intent intent = new Intent();
        intent.setAction("com.example.mybroadcastapplication.EXPLICIT_INTENT");
        getApplicationContext().sendBroadcast(intent);
    }

    @Override
    protected void onPostResume() {
        super.onPostResume();
        registerReceiver(myReceiver, intentFilter);
    }

    @Override
    protected void onStop() {
        super.onStop();
        unregisterReceiver(myReceiver);
    }

    @Override
    public void onClick(View v) {
        switch (v.getId()) {
            case R.id.button1:
                broadcastIntent();
                break;
            default:
        }
    }
}

我的 MyBroadcastReceiver.java 文件：

public class MyBroadcastReceiver extends BroadcastReceiver {

    @Override
    public void onReceive(Context context, Intent intent) {
        if (intent.getAction() != null && intent.getAction().equals("com.example.mybroadcastapplication.EXPLICIT_INTENT"))
            Toast.makeText(context, "Explicit intent received.", Toast.LENGTH_LONG).show();
    }
}

问候

Answer 1

如果您以编程方式注册/取消注册，则无需在清单中声明

BroadcastReceiver

。如果您希望从外部触发器（例如，在设备启动时或从警报管理器等）实例化它们，则只需在清单中声明 

BroadcastReceiver

Answer 2

从 API 级别 26 开始，您可以通过以下方式执行此操作：registerReceiver(some_broadreceiver, new IntentFilter("abcd"),Context.RECEIVER_NOT_EXPORTED);