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:当我不移动我的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; }
解决方法:
MOVE非常敏感,只要你的手指掉下来就会继续调用.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.
像这样的东西.
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; }
总结 以上是内存溢出为你收集整理的android – 即使没有移动,Motionevent Action_MOVE也会一直触发X和Y.全部内容,希望文章能够帮你解决android – 即使没有移动,Motionevent Action_MOVE也会一直触发X和Y.所遇到的程序开发问题。
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