
我有一个实现为WakefulIntentService的服务.每次启动适当的意图给负责的broadcastReceiver时,就会启动该服务.该服务在两种情况下启动:在设备启动时以及该服务的运行实例即将完成时,并通过要求AndroID的常规任务计划程序AlarmManager在将来的时间发出启动程序意图来计划新的执行.
问题是,出于安全原因,我have been advised不要在清单文件中声明的服务中使用androID:exported =“ true”.但是,如果忽略它,则会导致其中一部测试手机(运行AndroID 4.1.2的Samsung S3)拒绝执行服务:
06-13 11:34:34.181: W/ActivityManager(2270): Permission denIEd: checkComponentPermission() owningUID=1015506-13 11:34:34.181: W/ActivityManager(2270): Permission Denial: Accessing service ComponentInfo{com.mypackage.myapp/com.mypackage.myapp.MyService} from pID=10320, uID=2000 that is not exported from uID 10155添加androID:exported =“ true”可解决此问题.是否有其他选择可以避免执行被拒绝而不损害应用程序的安全性?
清单xml文件:
<?xml version="1.0" enCoding="utf-8"?><manifest xmlns:androID="http://schemas.androID.com/apk/res/androID" package="com.mypackage.myapp" androID:versionCode="3" androID:versionname="1.0"> <uses-sdk androID:minSdkVersion="8" androID:targetSdkVersion="17" /> <uses-permission androID:name="androID.permission.ACCESS_FINE_LOCATION" /> <uses-permission androID:name="androID.permission.INTERNET" /> <uses-permission androID:name="androID.permission.READ_PHONE_STATE" /> <uses-permission androID:name="androID.permission.RECEIVE_BOOT_COMPLETED" /> <uses-permission androID:name="androID.permission.WAKE_LOCK" /> <application androID:allowBackup="true" androID:icon="@drawable/ic" androID:label="@string/app_name" androID:theme="@style/Apptheme" > <receiver androID:name=".MybroadcastReceiver" > <intent-filter> <action androID:name="androID.intent.action.BOOT_COMPLETED" /> <category androID:name="androID.intent.category.DEFAulT" /> </intent-filter> <intent-filter> <action androID:name="com.mypackage.myapp" /> </intent-filter> </receiver> <service androID:name="com.mypackage.myapp.MyService" androID:exported="true"> </service> </application></manifest>broadcastReceiver:
package com.mypackage.myapp;import androID.content.broadcastReceiver;import androID.content.Context;import androID.content.Intent;import com.commonsware.cwac.wakeful.WakefulintentService;public class MybroadcastReceiver extends broadcastReceiver { @OverrIDe public voID onReceive(Context context, Intent intent) { WakefulintentService.senDWakefulWork(context, MyService.class); }}该服务在onDestroy()中包含启动程序意图调度代码:
public class MyService extends WakefulintentService {(...) @OverrIDe public voID doWakefulWork(Intent intent) { (...) } @OverrIDe public voID onDestroy() { PendingIntent pi = PendingIntent.getbroadcast(this, 0, new Intent("com.mypackage.myapp"), 0); AlarmManager am = (AlarmManager)(this.getSystemService(Context.ALARM_SERVICE)); am.set(AlarmManager.ELAPSED_REALTIME_WAKEUP, ONE_MINUTE, pi); super.onDestroy(); }}解决方法:
Actually it doesn’t occur from boot but when I attempt to start it manually on adb shell: am startservice com.mypackage.myapp/.MyService.
那不要那样做您的用户不会这样做.导出服务只是为了您可以运行adb shell命令,这并不是一个特别明智的举动.此外,您可以测试从adb shell发送启动时广播,达到相同的目的,而不必导出服务.
I haven’t been able to make the service start upon boot after removing the action string in my new Intent()
抱歉,我的意思是您的第二个动作字符串.您的<接收者>应该看起来像:
<receiver androID:name=".MybroadcastReceiver" > <intent-filter> <action androID:name="androID.intent.action.BOOT_COMPLETED" /> <category androID:name="androID.intent.category.DEFAulT" /> </intent-filter> </receiver>相应的PendingIntent将是:
PendingIntent pi = PendingIntent.getbroadcast(this, 0, new Intent(this, MybroadcastReceiver.class), 0); 总结 以上是内存溢出为你收集整理的Android服务执行被拒绝全部内容,希望文章能够帮你解决Android服务执行被拒绝所遇到的程序开发问题。
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