Question

我有一个应用程序，您应该可以使用我将在此问题中发布的代码完全且非常轻松地重新创建。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>

如您所见，它包含一个活动，一个（唤醒）广播接收器和一个intentservice，都在同一个包中。活动在发布时开始，这是代码：

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);
    }
}

这会成功触发onReceive的{{1}}功能。

TestReceiver

这是事情出错的地方，我在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); } }函数中放置了一个断点，它肯定会被调用。但是，onReceive类永远不会到达。我在MonitorService函数中放置了一个断点，但它似乎永远不会那么远。这是这个类的代码：

onHandleEvent

从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); } } }类的注释行可以看出，我尝试使用隐式意图而不是显式意图。我也读过this question并尝试了那里提到的一切。我在这里错过了什么吗？我在模拟器（Nexus7 API L）上运行它。

这里有什么我想念的吗？

Answer 1

Application Manifest中没有<intentservice>的标记。 IntentService是Service的子类，因此您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>

到

<service 
    android:name="com.example.broadcasttest.MonitorService"
    android:enabled="true" >
       <intent-filter>
           <action android:name="com.example.broadcasttest.MonitorService" />
       </intent-filter>
</service>

如何从WakefulBroadcastReceiver启动IntentService

1 个答案: