一个应用启动另外一个应用的Service

Posted 2022-01-31 zhangjin1120

应用A启动应用B的Service

• 应用A

public class MainActivity extends AppCompatActivity 

    @Override
    protected void onCreate(Bundle savedInstanceState) 
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

    

    public void startService(View view) 
        Intent intent = new Intent();
        intent.setComponent(new ComponentName("com.example.serviceb", "com.example.serviceb.AppService"));//第一个参数是包名，第二个是类名
        intent.putExtra("xx", "test");
        startService(intent);
    

• 被启动的应用B
  

        <service
            android:name=".AppService"
            android:exported="true" />

public class AppService extends Service 
    @Nullable
    @Override
    public IBinder onBind(Intent intent) 
        return null;
    

    @Override
    public int onStartCommand(Intent intent, int flags, int startId) 
        Log.i("xx", intent.getStringExtra("xx"));
        return super.onStartCommand(intent, flags, startId);
    

• 先把应用A，应用B都安装运行。点击应用A的按钮，就可以启动应用B的service了。如图：