如何启动电源按钮preSS应用程序 [英] how to start the app on power button press
问题描述
我要开始我的应用程序,当用户preSS的电源按钮。我米以下的<一个href="http://stackoverflow.com/questions/8940922/activate-an-application-when-a-power-button-is-clicked">This code
但它没有显示任何登录
和烤面包片。
I want to start my app when a user press the power button. I m following This code
but its not showing any Log
and toast.
这是我的完整的code。
here is my complete code.
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
}
}
menifest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;
}
}
- 我想我犯在我的
menifest文件中的错误
。请对此看看。谢谢。 - I think i m committing a mistake in my
menifest file
. please have a look on this. thanks.
推荐答案
首先,与其他广泛的铸造意图,为Intent.ACTION_SCREEN_OFF和Intent.ACTION_SCREEN_ON不能在你的Android清单申报他们!所以,你需要做,这将继续这样运行的服务
First, unlike other broad casted intents, for Intent.ACTION_SCREEN_OFF and Intent.ACTION_SCREEN_ON you CANNOT declare them in your Android Manifest! so You need to make a service which will keep on running like this
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
}
}
}
和你的接收器可以是一些
and your receiver can be something
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);
}
}
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