我有一个应用程序,您应该能够用我在这个问题中发布的代码完全轻松地重新创建它。这是清单文件:
<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.example.broadcasttest"
android:versionCode="1"
android:versionName="1.0" >
<uses-sdk
android:minSdkVersion="19"
android:targetSdkVersion="21" />
<uses-permission android:name="android.permission.WAKE_LOCK"/>
<application
android:allowBackup="true"
android:icon="@drawable/ic_launcher"
android:label="@string/app_name"
android:theme="@style/AppTheme" >
<activity
android:name="com.example.broadcasttest.MainActivity"
android:label="@string/app_name" >
<intent-filter>
<action android:name="android.intent.action.MAIN" />
<category android:name="android.intent.category.LAUNCHER" />
</intent-filter>
</activity>
<receiver
android:name="com.example.broadcasttest.TestReceiver"
android:label="@string/app_name"
android:enabled="true" >
</receiver>
<intentservice
android:name="com.example.broadcasttest.MonitorService"
android:enabled="true" >
<intent-filter>
<action android:name="com.example.broadcasttest.MonitorService" />
</intent-filter>
</intentservice>
</application>
</manifest>
正如您所看到的,包含一个活动、一个(唤醒)广播接收器和一个意图服务,所有这些都在同一个包中。活动在发布时开始,代码如下:
package com.example.broadcasttest;
import android.app.Activity;
import android.content.Intent;
import android.os.Bundle;
import android.view.Menu;
import android.view.MenuItem;
public class MainActivity extends Activity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_main);
sendBroadcast(new Intent(this, TestReceiver.class));
}
@Override
public boolean onCreateOptionsMenu(Menu menu) {
// Inflate the menu; this adds items to the action bar if it is present.
getMenuInflater().inflate(R.menu.main, menu);
return true;
}
@Override
public boolean onOptionsItemSelected(MenuItem item) {
// Handle action bar item clicks here. The action bar will
// automatically handle clicks on the Home/Up button, so long
// as you specify a parent activity in AndroidManifest.xml.
int id = item.getItemId();
if (id == R.id.action_settings) {
return true;
}
return super.onOptionsItemSelected(item);
}
}
这成功地触发了TestReceiver
的onReceive
功能。
package com.example.broadcasttest;
import android.content.Context;
import android.content.Intent;
import android.support.v4.content.WakefulBroadcastReceiver;
public class TestReceiver extends WakefulBroadcastReceiver {
@Override
public void onReceive(Context context, Intent intent) {
//Intent service = new Intent("com.example.broadcasttest.MonitorService");
Intent service = new Intent(context, MonitorService.class);
startWakefulService(context, service);
}
}
不过,这就是问题所在,我在onReceive
函数中放置了一个断点,它肯定会被调用。但是,MonitorService
类永远无法访问。我在onHandleEvent
函数中放置了一个断点,但它似乎从来没有这么远。这是这个类的代码:
package com.example.broadcasttest;
import android.app.IntentService;
import android.content.Intent;
public class MonitorService extends IntentService {
public MonitorService(String name) {
super(name);
}
public MonitorService()
{
super("MonitorService");
}
@Override
protected void onHandleIntent(Intent intent) {
try {
Thread.sleep(1000);
} catch (InterruptedException e) {
e.printStackTrace();
} finally {
TestReceiver.completeWakefulIntent(intent);
}
}
}
从TestReceiver
类中的注释行可以看出,我尝试使用隐式意图而不是显式意图。我也读过这个问题,并尝试了其中提到的所有内容。我是不是遗漏了什么?我正在一个模拟器(Nexus7 API L)上运行这个。
这里有我遗漏的东西吗?
应用程序清单中没有标记为<intentservice>
。IntentService
是Service
的一个子类,因此需要在清单中将其声明为服务。
更改
<intentservice
android:name="com.example.broadcasttest.MonitorService"
android:enabled="true" >
<intent-filter>
<action android:name="com.example.broadcasttest.MonitorService" />
</intent-filter>
</intentservice>
至
<service
android:name="com.example.broadcasttest.MonitorService"
android:enabled="true" >
<intent-filter>
<action android:name="com.example.broadcasttest.MonitorService" />
</intent-filter>
</service>