Question

我想根据接近传感器打开和关闭屏幕。我可以关闭屏幕。但屏幕背面的代码无效。有人可以帮我吗？这是代码：`

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);
      } 
}`

Answer 1

首先，我将屏幕亮度调暗尽可能低，然后使所有GUI元素都无法触摸以解决触摸问题。以下是我的代码：

@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，我参考了禁用整个屏幕视图的触摸。

Answer 2

要将屏幕亮度设置为on，值为1并将其设为off，则值为0.

在android中以编程方式打开和关闭屏幕

2 个答案: