Question

关于在启动时启动应用程序的 Stackoverflow 有很多问题。我对所有这些都进行了搜索，搜索谷歌但是它不起作用。

的AndroidManifest.xml

<uses-permission android:name="android.permission.RECEIVE_BOOT_COMPLETED"/>

<receiver
        android:name=".BootUpReceiver"
        android:enabled="true"
        android:exported="true"
        android:permission="android.permission.RECEIVE_BOOT_COMPLETED">
        <intent-filter>
            <action android:name="android.intent.action.BOOT_COMPLETED"/>
            <category android:name="android.intent.category.DEFAULT"/>
        </intent-filter>
</receiver>

BootUpReceiver

@Override
public void onReceive(Context context, Intent intent) {
    Log.w(TAG, "onReceive: starting app");
    Intent i = new Intent(context, Launcher.class);
    i.setFlags(Intent.FLAG_ACTIVITY_NEW_TASK);
    context.startActivity(i);
}

启动时的相关日志：

D/PackageManager: PROCAPK:/data/app/com.example.app-1  pkg name:com.example.app
W/PackageManager: Failure retrieving resources for com.example.app: Resource ID #0x0
AppOps: check op 53 uid 10109 for com.example.app

**BroadcastQueue: App ops not allowed for broadcast to uid 10109 pkg com.example.app**

D/BroadcastQueue: processNextBroadcast app ops skip = true, action = android.intent.action.BOOT_COMPLETED, package = com.example.app

I/LMS: find package:com.example.app, cert idx :19
                                 pkg key: long string

修改

此代码适用于模拟器。我想这确实取决于设备证券......任何想法？

在Boot上启动一个应用程序

0 个答案: