Question

我正在尝试简单的程序，如果移动有声音。所以在开始的时候我已经失声了 - 播放声音1然后每次移动它都会播放声音。在第4阶段，我从头开始玩。

问题在于：当我不移动手指并将其保持在同一位置时，声音仍然保持1比1 - 计算出x和y值的触发。我怎么阻止这个？

OnTouchListener MyOnTouchListener= new OnTouchListener()
{
public boolean onTouch(View view, MotionEvent event)
    {
    switch(event.getAction() & MotionEvent.ACTION_MASK)
        {

         case MotionEvent.ACTION_DOWN:
           x = (int) event.getX();
            y = (int) event.getY();    
            oldval = x+y;
            break;

    case MotionEvent.ACTION_MOVE:
        {
          Log.e("X value", "X is "+x);
          Log.e("Y value", "Y is "+y);
        try 
            {
                Thread.sleep(500);
            } catch (InterruptedException e) {
            }

            int newval= (int) (event.getX() + event.getY()); 

            if(Math.abs(oldval-newval)>50)
            {

                Log.e("First", "next button");
                longpressCount++;
                if(longpressCount==1)
                {
                   Log.e("1", "BUTTON PRESSED");
                }
                else if(longpressCount==2)
                {
                    Log.e("2", "BUTTON PRESSED");
                }
                else if(longpressCount==3)
                {
                    Log.e("3", "BUTTON PRESSED");
                }
                else if(longpressCount==4)
                {
                    Log.e("4", "BUTTON PRESSED");
                    longpressCount = 0;
                }

            }

            break;
         }
        }
        return true;
    }

Answer 1

MOVE是非常敏感的，只要你的手指向下，它就会继续被调用。在声音播放代码的末尾设置旧值，这样只有在距离该位置再移动50个距离时才会播放。

像这样。

OnTouchListener MyOnTouchListener= new OnTouchListener()
{
public boolean onTouch(View view, MotionEvent event)
    {
    switch(event.getAction() & MotionEvent.ACTION_MASK)
        {

         case MotionEvent.ACTION_DOWN:
           x = (int) event.getX();
            y = (int) event.getY();    
            oldval = x+y;
            break;

    case MotionEvent.ACTION_MOVE:
        {
          Log.e("X value", "X is "+x);
          Log.e("Y value", "Y is "+y);
        try 
            {
                Thread.sleep(500);
            } catch (InterruptedException e) {
            }

            int newval= (int) (event.getX() + event.getY()); 

            if(Math.abs(oldval-newval)>50)
            {

                Log.e("First", "next button");
                longpressCount++;
                if(longpressCount==1)
                {
                   Log.e("1", "BUTTON PRESSED");
                }
                else if(longpressCount==2)
                {
                    Log.e("2", "BUTTON PRESSED");
                }
                else if(longpressCount==3)
                {
                    Log.e("3", "BUTTON PRESSED");
                }
                else if(longpressCount==4)
                {
                    Log.e("4", "BUTTON PRESSED");
                    longpressCount = 0;
                }
                oldval = event.getX() + event.getY();
            }

            break;
         }
        }
        return true;
    }

即使没有移动，Motionevent Action_MOVE也会一直触发X和Y.

1 个答案: