Question

我正在使用绘图应用，onTouchEvents是标准的，并且想要添加Undo()函数来删除最后绘制的路径。

声明：

int thelastLineId=0;
private Bitmap bitmap; // drawing area for display or saving
private Canvas bitmapCanvas; // used to draw on bitmap
private Paint paintScreen; // use to draw bitmap onto screen
private Paint paintLine; // used to draw lines onto bitmap
private HashMap<Integer, Path> pathMap; // current Paths being drawn
private HashMap<Integer, Path> reservedpathMap; // for saving the paths being undone
private HashMap<Integer, Point> previousPointMap; // current Points

构造

pathMap = new HashMap<Integer, Path>();
reservedpathMap = new HashMap <Integer,Path>(); // for storing path being undone
previousPointMap = new HashMap<Integer, Point>();

的onDraw：

   @Override
   protected void onDraw(Canvas canvas) 
   {
       canvas.drawBitmap(bitmap, 0, 0, paintScreen);
       // for each path currently being drawn
       for (Integer key : pathMap.keySet()) 
            canvas.drawPath(pathMap.get(key), paintLine); // draw line     
   }

的onTouchEvent：

   @Override
   public boolean onTouchEvent(MotionEvent event) 
   {                  
      int action = event.getActionMasked(); // event type 
      int actionIndex = event.getActionIndex(); // pointer (i.e., finger)

      if (action == MotionEvent.ACTION_DOWN) 
      {       
          touchStarted(event.getX(actionIndex), event.getY(actionIndex), event.getPointerId(actionIndex));
      } 
      else if (action == MotionEvent.ACTION_UP) 
      {
          touchEnded(event.getPointerId(actionIndex));
      } 
      else 
      {
         touchMoved(event); 
      } 

      invalidate();
      return true; 
   }

touchStarted：

   private void touchStarted(float x, float y, int lineID) // lineID represents how many fingers, 1 finger 1 line
   {      
      Path path; // used to store the path for the given touch id
      Point point; // used to store the last point in path

      // if there is already a path for lineID
      if (pathMap.containsKey(lineID)) 
      {
         path = pathMap.get(lineID); // get the Path
         path.reset(); // reset the Path because a new touch has started
         point = previousPointMap.get(lineID); // get Path's last point
      } 
      else 
      {
         path = new Path(); // create a new Path
         pathMap.put(lineID, path); // add the Path to Map
         point = new Point(); // create a new Point
         previousPointMap.put(lineID, point); // add the Point to the Map
      } 

      path.moveTo(x, y);
      point.x = (int) x;  
      point.y = (int) y;  
   }

touchMoved：

   private void touchMoved(MotionEvent event) 
   {
      // for each of the pointers in the given MotionEvent
      for (int i = 0; i < event.getPointerCount(); i++) 
      {
         // get the pointer ID and pointer index
         int pointerID = event.getPointerId(i);
         int pointerIndex = event.findPointerIndex(pointerID);

         // if there is a path associated with the pointer
         if (pathMap.containsKey(pointerID)) 
         {
            float newX = event.getX(pointerIndex);
            float newY = event.getY(pointerIndex);

            // get the Path and previous Point associated with this pointer
            Path path = pathMap.get(pointerID);
            Point point = previousPointMap.get(pointerID);

            float deltaX = Math.abs(newX - point.x);
            float deltaY = Math.abs(newY - point.y);
            if (deltaX >= TOUCH_TOLERANCE || deltaY >= TOUCH_TOLERANCE) 
            {
               path.quadTo(point.x, point.y, ((newX + point.x)/2),((newY + point.y)/2));

               point.x = (int) newX ; 
               point.y = (int) newY ; 
         } 
      }      
   }

touchEnded：

   private void touchEnded(int lineID)
   {
      Path path = pathMap.get(lineID); // get the corresponding Path
      bitmapCanvas.drawPath(path, paintLine); 
      path.reset();           
   }

UNDO：

   public void undo()
   {
       Toast.makeText(getContext(), "undo button pressed" + thelastLineId, Toast.LENGTH_SHORT).show();

       Path path = pathMap.get(thelastLineId); 
       reservedpathMap.put(thelastLineId, path); // add the Path to reservedpathMap for later redo
       pathMap.remove(thelastLineId);   

       invalidate();          
   }

的问题：

我正在尝试使用上面显示的代码实现UNDO方法：尝试从thelastLindId HashMap中删除pathmap密钥（并将HashMap reservedpathMap添加到invalidate()稍后重做）这样当OnDraw()它将调用for (Integer key : pathMap.keySet()) canvas.drawPath(pathMap.get(key), paintLine);并重绘

{{1}}

然而，按下撤销按钮可以启动“撤消被点击”的祝酒，但最后绘制的行无法消失。

有人可以给我一个Undo（）和Redo（）的线索吗？非常感谢提前!!

Answer 1

我可以理解你想要实现的目标，你希望能够在画布上绘制线条，然后在项目中使用UNDO功能。首先，我认为当您向阵列添加路径时，应该是当用户抬起手指时，使用touchEnded方法。其次，我真的没有得到关于两个/三个手指的解释？你在画布上支持多点触控吗？这是我之前在一些示例中使用的实现，用于在具有撤消实现的画布上绘图。希望它能帮助你更清楚地做到：

public void onClickUndo () { 
if (paths.size()>0) { 
   undonePaths.add(paths.remove(paths.size()-1))
   invalidate();
 }
    else
     //toast the user 
}

public void onClickRedo (){
   if (undonePaths.size()>0) { 
       paths.add(undonePaths.remove(undonePaths.size()-1)) 
       invalidate();
   } 
   else 
     //toast the user 
}

以下是触摸方法的等价物：

@Override
protected void onSizeChanged(int w, int h, int oldw, int oldh) {
    super.onSizeChanged(w, h, oldw, oldh);
}

@Override
protected void onDraw(Canvas canvas) {            

    for (Path p : paths){
        canvas.drawPath(p, mPaint);
    }

}

private float mX, mY;
private static final float TOUCH_TOLERANCE = 0;

private void touch_start(float x, float y) {
    mPath.reset();
    mPath.moveTo(x, y);
    mX = x;
    mY = y;
}
private void touch_move(float x, float y) {
    float dx = Math.abs(x - mX);
    float dy = Math.abs(y - mY);
    if (dx >= TOUCH_TOLERANCE || dy >= TOUCH_TOLERANCE) {
        mPath.quadTo(mX, mY, (x + mX)/2, (y + mY)/2);
        mX = x;
        mY = y;
    }
}
private void touch_up() {
    mPath.lineTo(mX, mY);
    // commit the path to our offscreen
    mCanvas.drawPath(mPath, mPaint);
    // kill this so we don't double draw            
    mPath = new Path();
    paths.add(mPath);
}

从这里你可以看到路径是我存储路径的arraylist。如果你告诉我为什么你需要把你的路径放在hashmap中，我可能会更有帮助。

android（多点触控）绘图app撤消功能

的问题：

1 个答案: