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我正在开发一个绘图应用程序,它onTouchEvents是标准的,并且想添加Undo()功能以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

构造函数:

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

绘制:

   @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     
   } 

触摸事件:

   @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; 
   } 

触摸开始:

   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;  
   } 

触摸移动:

   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 ; 
         } 
      }      
   } 

触摸结束:

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

撤消:

   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();          
   } 

问题:

我正在尝试使用如上所示的代码来实现 UNDO 方法:尝试从(并放入以供以后重做)中删除thelastLindId密钥,以便它何时调用和重绘HashMap pathmapHashMap reservedpathMapinvalidate()OnDraw()

for (Integer key : pathMap.keySet()) 
                canvas.drawPath(pathMap.get(key), paintLine); 

但是,按下撤消按钮可以启动“单击撤消”的吐司,但最后绘制的线未能消失。

有人可以给我一个 Undo() 和 Redo() 的线索吗?提前谢谢了!!

4

1 回答 1

1

我可以理解您想要实现的目标,您希望能够在画布上绘制线条,然后为您的项目提供 UNDO 功能。touchEnded首先,我认为将路径添加到数组的那一刻应该是用户在您的方法中抬起手指的时候。其次,我真的不明白你解释的关于两个/三个手指的事情?你的画布支持多点触控吗?这是我之前在某些示例中使用的实现,用于在具有撤消实现的画布上绘图。希望它可以帮助您使事情更清楚:

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 
}

以下是您的触摸方法的等价物:

@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);
}

从这里可以看出,paths 是我存储路径的数组列表。如果您告诉我为什么需要将路径放在 hashmap 中,也许我会更有帮助。

于 2013-02-18T19:19:37.940 回答