android - 如何在按下电源按钮时启动应用程序

Question

我想在用户按下电源按钮时启动我的应用程序。我正在关注 This code但它没有显示任何 Log 和 toast。

这是我的完整代码。

MyReceiver.java

import android.content.BroadcastReceiver;
   import android.content.Context;
   import android.content.Intent;
   import android.util.Log;
   import android.widget.Toast;

   public class MyReceiver extends BroadcastReceiver {

@Override
public void onReceive(Context context, Intent intent) {

    // TODO Auto-generated method stub

    Log.v("onReceive", "Power button is pressed.");

    Toast.makeText(context, "power button clicked", Toast.LENGTH_LONG)
            .show();

    // perform what you want here

}

}

list 文件

<?xml version="1.0" encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
package="com.example.powerbuttontest"
android:versionCode="1"
android:versionName="1.0" >

<uses-sdk
    android:minSdkVersion="8"
    android:targetSdkVersion="17" />

<application
    android:allowBackup="true"
    android:icon="@drawable/ic_launcher"
    android:label="@string/app_name"
    android:theme="@style/AppTheme" >
    <activity
        android:name="com.example.powerbuttontest.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=".MyReceiver" >
        <intent-filter>
            <action android:name="android.intent.action.SCREEN_OFF" >
            </action>
            <action android:name="android.intent.action.SCREEN_ON" >
            </action>
            <action android:name="android.intent.action.ACTION_POWER_CONNECTED" >
            </action>
            <action android:name="android.intent.action.ACTION_POWER_DISCONNECTED" >
            </action>
            <action android:name="android.intent.action.ACTION_SHUTDOWN" >
            </action>
        </intent-filter>
    </receiver>
</application>
</manifest>

MainActivity.java

package com.example.powerbuttontest;

import android.os.Bundle;
import android.app.Activity;
import android.view.Menu;

public class MainActivity extends Activity {

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

@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;
}

}

• 我想我在我的menifest 文件 中犯了一个错误。请看看这个。谢谢。

Answer 1

首先，与其他广泛的 Intent 不同，对于 Intent.ACTION_SCREEN_OFF 和 Intent.ACTION_SCREEN_ON，您不能在 Android list 中声明它们!所以你需要做一个像这样继续运行的服务

public static class UpdateService extends Service {

        @Override
        public void onCreate() {
            super.onCreate();
            // register receiver that handles screen on and screen off logic
            IntentFilter filter = new IntentFilter(Intent.ACTION_SCREEN_ON);
            filter.addAction(Intent.ACTION_SCREEN_OFF);
            BroadcastReceiver mReceiver = new Receiver();
            registerReceiver(mReceiver, filter);
        }

        @Override
        public void onStart(Intent intent, int startId) {
            boolean screenOn = intent.getBooleanExtra("screen_state", false);
            if (!screenOn) {
                // your code
            } else {
                // your code
            }
        }
}

你的接收器可以是什么东西

public class Receiver extends BroadcastReceiver {

    private boolean screenOff;

    @Override
    public void onReceive(Context context, Intent intent) {
        if (intent.getAction().equals(Intent.ACTION_SCREEN_OFF)) {
            screenOff = true;
        } else if (intent.getAction().equals(Intent.ACTION_SCREEN_ON)) {
            screenOff = false;
        }
        Intent i = new Intent(context, UpdateService.class);
        i.putExtra("screen_state", screenOff);
        context.startService(i);
    }

}