查看不显示onTouch

Question

我正在开发一个应用程序，我想在触摸事件注册时显示视图。问题是视图不会显示，我的代码变得没有反应。

这是我的观点：

public class TouchFX extends View { 

RectF oval; 
float Tx, Ty; 
Paint paint; 
int longSide, numRun; 
float top, left, right, bottom; 
boolean removable; 

public TouchFX(Context context) { 
    super(context); 
    // TODO Auto-generated constructor stub 
    Tx = TouchService.x; 
    Ty = TouchService.y; 
    paint = new Paint(); 
    paint.setColor(Color.WHITE); 
    paint.setAntiAlias(true); 
    paint.setStrokeJoin(Paint.Join.ROUND); 
    numRun = 0; 
    removable = false; 
} 

@Override 
protected void onDraw(Canvas canvas) { 
    // TODO Auto-generated method stub 
    super.onDraw(canvas); 

    if (canvas.getWidth() > canvas.getHeight()) 
     longSide = canvas.getWidth(); 
    else 
     longSide = canvas.getHeight(); 

    if (numRun == 0) { 
     left = Tx - (longSide/15); 
     top = Ty + (longSide/15); 
     right = Tx + (longSide/15); 
     bottom = Ty - (longSide/15); 
    } else if (numRun <= 8) { 
     left += 2; 
     top += 2; 
     right += 2; 
     bottom += 2; 
    } 
     oval = new RectF(left, top, right, bottom); 

    if (numRun < 9) 
     canvas.drawArc(oval, 0, 360, false, paint); 
    else 
     removable = true; 

    numRun++; 

    try { 
     Thread.sleep(400); 
    } catch (InterruptedException e) { 
     // TODO Auto-generated catch block 
     e.printStackTrace(); 
     } 
    } 
} 

这里是我的onTouch处理程序：

public boolean onTouch(View v, MotionEvent event) { 
    // TODO Auto-generated method stub 
    if (v == tv && event.getAction() == MotionEvent.ACTION_DOWN 
      && isRunning == true) { 
     x = event.getX(); 
     y = event.getY(); 

     tfx = new TouchFX(this); 

     winmngr.addView(tfx, lpFX); 


        //Do some stuff, make some Toasts 


     while(!tfx.removable) { 
      //waiting for animation to end 
     } 

     winmngr.removeView(tfx); 
    } 
    return false; 
} 

现在，很明显它被陷在while循环中，但为什么不是绘制弧形和更新布尔值？

Answer 1

它卡住，因为onTouch事件在UI线程（这样的onDraw）工作。因此，您在while循环中阻止您的UI线程，并且onDraw（）不会获得处理器时间。

考虑使用回调机制，因此removeView()例程由您的onDraw方法触发。 UI线程中的sleep(400)也是非常糟糕的主意。您可以尝试使用处理程序并执行延迟执行：

public class YouCustomView extends View 
{ 

    Handler handler = new Handler(); 

    @Override 
    protected onDraw(Canvas canvas){ 
     //your draw routine here 
     .... 

     //wait 400 ms and remove view 
     handler.postDelayed(new Runnable() 
     { 
      @Override 
      public void run() 
      { 
       //trigger callback to remove view 
      } 
     }, 400); 

Answer 2

这里是我的代码为我完全地工作，就会得出一个字符串，希望它会帮助你解决问题

@Override 
public boolean onTouchEvent(MotionEvent event) { 
    // draw text at touch 
    try { 
     canvas = getHolder().lockCanvas(); 
     synchronized (getHolder()) { 
      if (event.getAction() == MotionEvent.ACTION_DOWN 
        || event.getAction() == MotionEvent.ACTION_MOVE) { 
       // clear the screen 
       canvas.drawColor(Color.BLACK); 
       // draw glyphs 
       glyphs.drawString(canvas, "Drawing string at " 
         + (int) event.getX() + " " + (int) event.getY(), 
         (int) event.getX(), (int) event.getY()); 
      } 
      if (event.getAction() == MotionEvent.ACTION_UP) { 
       canvas.drawColor(Color.BLACK); 
       glyphs.drawString(canvas, "Drawn string at " 
         + (int) event.getX() + " " + (int) event.getY(), 
         (int) event.getX(), (int) event.getY()); 
      } 
     } 
    } finally { 
     if (canvas != null) { 
      getHolder().unlockCanvasAndPost(canvas); 
     } 
    } 
    // event was handled 
    return true; 
}