打开屏幕上，并关闭在编程机器人 [英] Turn screen on and off programatically in android

问题描述

我想根据接近传感器打开屏幕开和关。我能够关闭屏幕。但code到屏幕背面是行不通的。任何人都可以帮助我吗？
这是code：`

 公共无效onSensorChanged（SensorEvent事件）{
如果（event.values​​ [0] == 0）{Toast.makeText（getApplicationContext（），传感器中0，Toast.LENGTH_LONG）.show（）;
WindowManager.LayoutParams PARAMS = getWindow（）的getAttributes（）。
params.flags | = LayoutParams.FLAG_KEEP_SCREEN_ON;
params.screenBrightness = 0;
。getWindow（）setAttributes（PARAMS）;
      }其他{Toast.makeText（getApplicationContext（），传感器1，Toast.LENGTH_LONG）.show（）;
WindowManager.LayoutParams PARAMS = getWindow（）的getAttributes（）。params.screenBrightness = -1;
。getWindow（）setAttributes（PARAMS）;
      }
}`
 

解决方案

首先，我调暗屏幕亮度尽可能低的水平，然后作出的所有GUI元素不可点击触摸的问题。以下是我的code：

  @覆盖
公共无效onSensorChanged（SensorEvent事件）{
    // TODO自动生成方法存根
    WindowManager.LayoutParams PARAMS = this.getWindow（）的getAttributes（）。    如果（event.values​​ [0] == 0）{
        // TODO商店原来的亮度值
        params.screenBrightness = 0.005f;
        。this.getWindow（）setAttributes（PARAMS）;
        enableDisableViewGroup（（ViewGroup中）findViewById（R.id.YOUR_MAIN_LAYOUT）.getParent（），FALSE）;
        Log.e（onSensorChanged，近）;    }其他{
        // TODO商店原来的亮度值
        params.screenBrightness = -1.0F;
        。this.getWindow（）setAttributes（PARAMS）;
        enableDisableViewGroup（（ViewGroup中）findViewById（R.id.YOUR_MAIN_LAYOUT）.getParent（），TRUE）;
        Log.e（onSensorChanged，远）;
    }
}
 

从这里，我参考了禁用触控对于整个屏幕的看法。

I want to turn screen ON and OFF based on the proximity sensor. I am able to turn the screen off. but the code to ON the screen back is not working. Can anyone help me please? This is the code:`

public void onSensorChanged(SensorEvent event) {
if (event.values[0] == 0) {

Toast.makeText(getApplicationContext(), "sensor in 0",Toast.LENGTH_LONG).show();
WindowManager.LayoutParams params = getWindow().getAttributes();
params.flags |= LayoutParams.FLAG_KEEP_SCREEN_ON;
params.screenBrightness = 0;
getWindow().setAttributes(params);


      } else {

Toast.makeText(getApplicationContext(), "sensor in 1",Toast.LENGTH_LONG).show();
WindowManager.LayoutParams params = getWindow().getAttributes();

params.screenBrightness = -1;
getWindow().setAttributes(params);
      } 
}`

解决方案

First I dimmed screen brightness as low as possible and then made all GUI elements unclickable for touch issues. Following is my code:

@Override
public void onSensorChanged(SensorEvent event) {
    // TODO Auto-generated method stub  
    WindowManager.LayoutParams params = this.getWindow().getAttributes();

    if (event.values[0] == 0) {
        //TODO Store original brightness value
        params.screenBrightness = 0.005f;
        this.getWindow().setAttributes(params);
        enableDisableViewGroup((ViewGroup)findViewById(R.id.YOUR_MAIN_LAYOUT).getParent(),false);
        Log.e("onSensorChanged","NEAR");

    } else {
        //TODO Store original brightness value          
        params.screenBrightness = -1.0f;
        this.getWindow().setAttributes(params);                     
        enableDisableViewGroup((ViewGroup)findViewById(R.id.YOUR_MAIN_LAYOUT).getParent(),true);
        Log.e("onSensorChanged","FAR");  
    }       
}  

From here, I took reference to disable touch for entire screen's view.

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