0

这是我的服务班

    import android.app.Service;
    import android.content.Intent;
    import android.os.IBinder;
    import android.util.Log;

    public class Communicationservice extends Service {

        @Override
        public IBinder onBind(Intent intent) {
            // TODO Auto-generated method stub
            return null;
        }

        @Override
        public void onCreate() {
            super.onCreate();
            Log.d("I AM Service","Service Created");
        }

        @Override
        public void onStart(Intent intent, int startId) {
            // TODO Auto-generated method stub
            super.onStart(intent, startId);
        }

    }

Here is my Main Activity


import android.os.Bundle;
import android.app.Activity;
import android.content.Intent;
import android.util.Log;
import android.view.Menu;

public class MainActivity extends Activity {

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


        Log.d("flow","1");
        Intent in = new Intent("com.yal.Communication.Communicationservice");
        this.startService(in);
        //startService(new Intent(MainActivity.this,Communicationservice.class));
        Log.d("flow","2");
    }

    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        getMenuInflater().inflate(R.menu.activity_main, menu);
        return true;
    }
}

这是 Mainfest.xml

<application
        android:icon="@drawable/ic_launcher"
        android:label="@string/app_name"
        android:theme="@style/AppTheme" >

        <activity > ...  </activity>
        <service android:enabled="true" android:name="com.yal.Communication.Communicationservice"></service>
    </application>

错误

11-02 16:34:17.939: D/flow(1390): 1
11-02 16:34:17.939: W/ActivityManager(92): Unable to start service Intent { act=com.yal.Communication.Communicationservice }: not found
11-02 16:34:17.939: D/flow(1390): 2
4

2 回答 2

2

你需要检查意图过滤器。

下面的片段会帮助你。

<service android:name="com.yal.Communication.Communicationservice">
    <intent-filter>
        <action android:name="com.yal.Communication.Communicationservice" />
    </intent-filter>
</service>

然后你可以像下面这样调用这个服务。

Intent serviceIntent = new Intent("com.yal.Communication.Communicationservice");
startService(serviceIntent);
于 2012-11-02T11:21:02.463 回答
0

构造函数public Intent (String action)可用于指定操作，请参见以下 java doc：

public Intent (String action)

Create an intent with a given action. All other fields (data, type, class) are null. Note that the action must be in a namespace because Intents are used globally in the system -- for example the system VIEW action is android.intent.action.VIEW; an application's custom action would be something like com.google.app.myapp.CUSTOM_ACTION.

Parameters
action  The Intent action, such as ACTION_VIEW.

对您的意图创建进行以下更改：

Intent in = new Intent(this,com.yal.Communication.Communicationservice.class);

或者在 me 清单文件中指定操作：

<service android:name="com.yal.Communication.Communicationservice">
    <intent-filter>
        <action android:name="com.yal.Communication.Communicationservice" />
    </intent-filter>
</service>
于 2012-11-02T11:21:57.160 回答