我从一个询问如何在Android中绘制的问题中得到了这个代码的答案,但是在使用它并在我的应用程序中测试时,我发现在绘制大件事物或许多路径时效率不高.问题来自onDraw中的代码,因为每次调用invalidate()时都会调用onDraw,其中包含一个循环,它将所有路径再次绘制到画布,并且通过向其添加更多路径,它变得非常慢.
这是班级：
public class DrawingView extends View implements OnTouchListener {
private Canvas m_Canvas;
private Path m_Path;
private Paint m_Paint;
ArrayList<Pair<Path, Paint>> paths = new ArrayList<Pair<Path, Paint>>();
ArrayList<Pair<Path, Paint>> undonePaths = new ArrayList<Pair<Path, Paint>>();
private float mX, mY;
private static final float TOUCH_TOLERANCE = 4;
public static boolean isEraserActive = false;
private int color = Color.BLACK;
private int stroke = 6;
public DrawingView(Context context, AttributeSet attr) {
super(context);
setFocusable(true);
setFocusableInTouchMode(true);
setBackgroundColor(Color.WHITE);
this.setOnTouchListener(this);
onCanvasInitialization();
}
public void onCanvasInitialization() {
m_Paint = new Paint();
m_Paint.setAntiAlias(true);
m_Paint.setDither(true);
m_Paint.setColor(Color.parseColor("#000000"));
m_Paint.setStyle(Paint.Style.STROKE);
m_Paint.setStrokeJoin(Paint.Join.ROUND);
m_Paint.setStrokeCap(Paint.Cap.ROUND);
m_Paint.setStrokeWidth(2);
m_Canvas = new Canvas();
m_Path = new Path();
Paint newPaint = new Paint(m_Paint);
paths.add(new Pair<Path, Paint>(m_Path, newPaint));
}
@Override
public void setBackground(Drawable background) {
mBackground = background;
super.setBackground(background);
}
@Override
protected void onSizeChanged(int w, int h, int oldw, int oldh) {
super.onSizeChanged(w, h, oldw, oldh);
}
public boolean onTouch(View arg0, MotionEvent event) {
float x = event.getX();
float y = event.getY();
switch (event.getAction()) {
case MotionEvent.ACTION_DOWN:
touch_start(x, y);
invalidate();
break;
case MotionEvent.ACTION_MOVE:
touch_move(x, y);
invalidate();
break;
case MotionEvent.ACTION_UP:
touch_up();
invalidate();
break;
}
return true;
}
@Override
protected void onDraw(Canvas canvas) {
for (Pair<Path, Paint> p : paths) {
canvas.drawPath(p.first, p.second);
}
}
private void touch_start(float x, float y) {
if (isEraserActive) {
m_Paint.setColor(Color.WHITE);
m_Paint.setStrokeWidth(50);
Paint newPaint = new Paint(m_Paint); // Clones the mPaint object
paths.add(new Pair<Path, Paint>(m_Path, newPaint));
} else {
m_Paint.setColor(color);
m_Paint.setStrokeWidth(stroke);
Paint newPaint = new Paint(m_Paint); // Clones the mPaint object
paths.add(new Pair<Path, Paint>(m_Path, newPaint));
}
m_Path.reset();
m_Path.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) {
m_Path.quadTo(mX, mY, (x + mX) / 2, (y + mY) / 2);
mX = x;
mY = y;
}
}
private void touch_up() {
m_Path.lineTo(mX, mY);
// commit the path to our offscreen
m_Canvas.drawPath(m_Path, m_Paint);
// kill this so we don't double draw
m_Path = new Path();
Paint newPaint = new Paint(m_Paint); // Clones the mPaint object
paths.add(new Pair<Path, Paint>(m_Path, newPaint));
}
public void onClickUndo() {
if (!paths.isEmpty()) {//paths.size() > 0) {
undonePaths.add(paths.remove(paths.size() - 1));
undo = true;
invalidate();
}
}
public void onClickRedo() {
if (!undonePaths.isEmpty()){//undonePaths.size() > 0) {
paths.add(undonePaths.remove(undonePaths.size() - 1));
undo = true;
invalidate();
}
}}
但我再次在互联网上搜索以找到更好的绘图方式,所以我找到了以下内容：
1将以下内容添加到构造函数中：
mBitmapPaint = new Paint(Paint.DITHER_FLAG);
2使用以下代码覆盖onSizeChanged：
protected void onSizeChanged(int w, int h, int oldw, int oldh) {
super.onSizeChanged(w, h, oldw, oldh);
mBitmap = Bitmap.createBitmap(w, h, Bitmap.Config.ARGB_4444);
m_Canvas = new Canvas(mBitmap);
}
3把它放在onDraw中：
protected void onDraw(Canvas canvas) {
canvas.drawBitmap(mBitmap, 0, 0, mBitmapPaint);
if (!paths.isEmpty())
canvas.drawPath(paths.get(paths.size() - 1).first, paths.get(paths.size() - 1).second);
}
这种方法有效并且不会减慢视图速度,但这种方法的问题在于我无法使用撤消和重做功能.
我尝试了很多东西来做第二种方法的撤销和重做,但是我无法做到.所以我在这里问的是三件事之一：
1.使用第二种方法进行撤消和重做的方法
2.另一种可以进行撤销和重做的方法
一个全新的类,已经完成了所有工作,比如一个开源库或其他东西.
如果可以的话请帮忙.
谢谢
编辑1
好吧,所以我把它限制在此,然后我再做不了什么,我已经尝试了8个多小时了.它一直工作直到撤消(你可以撤消任意数量的路径),然后当再次绘制所有剩余的路径消失时,我不知道是什么让它做到这一点.
@Override
protected void onDraw(Canvas canvas) {
if (mBitmap != null)
canvas.drawBitmap(mBitmap, 0, 0, mBitmapPaint);
if (!paths.isEmpty() && !undo)
canvas.drawPath(paths.get(paths.size() - 1).first, paths.get(paths.size() - 1).second);
if (undo) {
setBackground(mBackground);
for (Pair<Path, Paint> p : paths)
canvas.drawPath(p.first, p.second);
mBitmap = Bitmap.createBitmap(w, h, Bitmap.Config.ARGB_4444);
m_Canvas = new Canvas(mBitmap);
undo = false;
}
}
基本上我所做的就是首先使用第一种方法(在调用undo之前),然后如果单击undo,则将undo设置为true,并执行if(undo)下的代码,这实际上是第一种方法(计算所有路径)再次),然后我将计算所有路径的结果再次绘制到mBitmap中,因此每当再次调用onDraw时,它会在此之上绘制,但该部分仍然需要工作,我希望有人可以帮助这部分.
解决方法:
处理这种情况的方法是使用具有视图大小的Bitmap.在触摸事件上,绘制到位图的画布.在onDraw中,只需将位图绘制到0,0的画布上.对于撤消/重做,.你可以擦除位图并重新绘制所有路径.这需要花费更长的时间,但每次撤消/重做只发生一次.
如果用户通常执行一次撤消/重做.你可以通过另一个位图退一步来优化.