14

我正在尝试在 android 中的单击方法上绘制标记。当我绘制标记时，它会绘制，但绘制需要更多时间，即 30-40 毫秒，有时需要 2-3 秒。这是我有draw方法的类的代码。

public class MyItemizedOverlay extends ItemizedOverlay<OverlayItem> {

    private ArrayList<OverlayItem> overlayItemList = new ArrayList<OverlayItem>();

    public MyItemizedOverlay(Drawable pDefaultMarker,
            ResourceProxy pResourceProxy) {
        super(pDefaultMarker, pResourceProxy);
    }

    @Override
    public void draw(Canvas canvas, MapView mapView, boolean arg2) {
        super.draw(canvas, mapView, arg2);

        // ---translate the GeoPoint to screen pixels---
        Point screenPts = new Point();
        mapView.getProjection().toPixels(p, screenPts);

        // ---add the marker---
        Bitmap bmp = BitmapFactory.decodeResource(getResources(),
                R.drawable.pin_annotation_darkblue);
        Bitmap bmp1 = BitmapFactory.decodeResource(getResources(),
                R.drawable.pin_annotation_green);
        Bitmap bmp2 = BitmapFactory.decodeResource(getResources(),
                R.drawable.pin_annotation_bue);
        Bitmap bmp3 = BitmapFactory.decodeResource(getResources(),
                R.drawable.pin_annotation_light);
        Bitmap bmp4 = BitmapFactory.decodeResource(getResources(),
                R.drawable.pin_annotation_light);
        Bitmap bmp5 = BitmapFactory.decodeResource(getResources(),
                R.drawable.pin_annotation_light);
        Bitmap bmp6 = BitmapFactory.decodeResource(getResources(),
                R.drawable.pin_annotation_light);
        if (count == 1) {
            int caller = getIntent().getIntExtra("button", 0);
            switch (caller) {
            case R.id.btMap:
                canvas.drawBitmap(bmp, screenPts.x, screenPts.y - 50, null);
                bmp.recycle();
                break;
            case R.id.imageButton1:
                canvas.drawBitmap(bmp1, screenPts.x, screenPts.y - 50, null);
                bmp1.recycle();
                break;
            case R.id.imageButton2:
                canvas.drawBitmap(bmp2, screenPts.x, screenPts.y - 50, null);
                bmp2.recycle();
                break;
            case R.id.imageButton3:
                canvas.drawBitmap(bmp3, screenPts.x, screenPts.y - 50, null);
                bmp3.recycle();
                break;
            case R.id.imageButton4:
                canvas.drawBitmap(bmp4, screenPts.x, screenPts.y - 50, null);
                bmp4.recycle();
                break;
            case R.id.imageButton5:
                canvas.drawBitmap(bmp5, screenPts.x, screenPts.y - 50, null);
                bmp5.recycle();
                break;
            case R.id.imageButton6:
                canvas.drawBitmap(bmp6, screenPts.x, screenPts.y - 50, null);
                bmp6.recycle();
                break;
            }
        }
        // Bitmap bmp = BitmapFactory.decodeResource(getResources(),
        // R.drawable.pin_annotation_green);
        // if (count == 1) {
        // canvas.drawBitmap(bmp, screenPts.x, screenPts.y - 50, null);
        // }
}
4

3 回答 3

20

您应该在构造函数中初始化所有位图。解码位图需要很长时间。您可以使用HashMap(key, value) 来存储它们。然后在onDraw中，获取匹配的位图，直接绘制。

例如

public class MyView extends View{

    private HashMap<String, Bitmap> mStore = new HashMap<String, Bitmap>();
    public MyView(Context context) {
        super(context);
        // TODO Auto-generated constructor stub

        init();
    }

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

        int caller = getIntent().getIntExtra("button", 0);
        Bitmap bmp = null;
        switch (caller) {
        case R.id.btMap:
            bmp = mStore.get(R.id.btMap);
            canvas.drawBitmap(bmp, screenPts.x, screenPts.y - 50, null);
            bmp.recycle();
            bmp = null;
            break;
        case R.id.imageButton1:
            bmp = mStore.get(R.id.imageButton1);
            canvas.drawBitmap(bmp, screenPts.x, screenPts.y - 50, null);
            bmp1.recycle();
            bmp1 = null;
            break;
        }

        super.onDraw(canvas);
    }

    public void init() {
        Bitmap bmp = BitmapFactory.decodeResource(getResources(),
                R.drawable.pin_annotation_darkblue);
        mStore.put(R.id.btMap, bmp);

        bmp = BitmapFactory.decodeResource(getResources(),
                R.drawable.pin_annotation_green);
        mStore.put(R.id.imageButton1, bmp);
    }
}

这是我根据您的代码所做的。您必须检查一些重复的资源 ID。

private ArrayList<OverlayItem> overlayItemList = new ArrayList<OverlayItem>();
private HashMap<String, Bitmap> mStore = new HashMap<String, Bitmap>();

public MyItemizedOverlay(Drawable pDefaultMarker,
        ResourceProxy pResourceProxy) {
    super(pDefaultMarker, pResourceProxy);

    Bitmap bmp = BitmapFactory.decodeResource(getResources(),
            R.drawable.pin_annotation_darkblue);
    mStore.put(R.id.btMap, bmp);
    bmp = BitmapFactory.decodeResource(getResources(),
            R.drawable.pin_annotation_green);
    mStore.put(R.id.imageButton1, bmp);
    bmp = BitmapFactory.decodeResource(getResources(),
            R.drawable.pin_annotation_bue);
    mStore.put(R.id.imageButton2, bmp);
    bmp = BitmapFactory.decodeResource(getResources(),
            R.drawable.pin_annotation_light); 
    mStore.put(R.id.imageButton3, bmp);
    bmp = BitmapFactory.decodeResource(getResources(),
            R.drawable.pin_annotation_light); // check it
    mStore.put(R.id.imageButton4, bmp);
    bmp = BitmapFactory.decodeResource(getResources(),
            R.drawable.pin_annotation_light); // check it
    mStore.put(R.id.imageButton5, bmp);
    bmp = BitmapFactory.decodeResource(getResources(),
            R.drawable.pin_annotation_light); // check it
    mStore.put(R.id.imageButton6, bmp);

}

@Override
public void draw(Canvas canvas, MapView mapView, boolean arg2) {
    super.draw(canvas, mapView, arg2);

    // ---translate the GeoPoint to screen pixels---
    Point screenPts = new Point();
    mapView.getProjection().toPixels(p, screenPts);

    // ---add the marker---
    if (count == 1) {
        int caller = getIntent().getIntExtra("button", 0);
        Bitmap bmp = null;

        switch (caller) {
        case R.id.btMap:
            bmp = mStore.get(R.id.btMap);
            canvas.drawBitmap(bmp, screenPts.x, screenPts.y - 50, null);
            bmp.recycle();
            break;
        case R.id.imageButton1:
            bmp = mStore.get(R.id.imageButton1);
            canvas.drawBitmap(bmp1, screenPts.x, screenPts.y - 50, null);
            bmp.recycle();
            break;
        case R.id.imageButton2:
            bmp = mStore.get(R.id.imageButton2);
            canvas.drawBitmap(bmp, screenPts.x, screenPts.y - 50, null);
            bmp.recycle();
            break;
        case R.id.imageButton3:
            bmp = mStore.get(R.id.imageButton3);
            canvas.drawBitmap(bmp, screenPts.x, screenPts.y - 50, null);
            bmp.recycle();
            break;
        case R.id.imageButton4:
            bmp = mStore.get(R.id.imageButton4);
            canvas.drawBitmap(bmp, screenPts.x, screenPts.y - 50, null);
            bmp.recycle();
            break;
        case R.id.imageButton5:
            bmp = mStore.get(R.id.imageButton5);
            canvas.drawBitmap(bmp, screenPts.x, screenPts.y - 50, null);
            bmp.recycle();
            break;
        case R.id.imageButton6:
            bmp = mStore.get(R.id.imageButton6);
            canvas.drawBitmap(bmp, screenPts.x, screenPts.y - 50, null);
            bmp.recycle();
            break;
        }
    }
    // Bitmap bmp = BitmapFactory.decodeResource(getResources(),
    // R.drawable.pin_annotation_green);
    // if (count == 1) {
    // canvas.drawBitmap(bmp, screenPts.x, screenPts.y - 50, null);
    // }
}
于 2013-10-28T07:05:30.960 回答
8

优化代码的想法是只执行绘图所需的操作。所以你，应该从你的 onDraw 方法中删除：

  • 任何实例化：它们需要很长时间，经常调用 onDraw 并且您不想创建这么多新对象。在 onLayout 期间存储 screenPts 并始终重复使用相同的点。
  • BitmapFactory.decodeResource ：这需要相当长的时间。首先解码您的位图，存储它们并且仅在 onDraw 期间绘制它们。
  • 当您不再需要它们时回收位图，而不是每次绘制它们时。

例如 ：

  • 在 onResume 期间解码您的位图
  • 在 onPause 期间回收它们
  • 解码应该发生在异步任务中。当异步任务结束时，引发一个标志以向 onDraw 指示图像已准备好并且可以绘制。
  • 在后台解码图像非常重要，因为它需要很长时间。不要在主 UI 线程中执行此操作。否则你的应用会看起来没有响应
  • 在 onLayout 中计算你的 screenPts 并一直重复使用相同的点。
  • 也不要在 onDraw 期间调用 getIntent。

简而言之，在 onDraw 期间最小化操作，您将实现非常快的绘图，大约 60 FPS。

您还应该考虑删除那个（丑陋的）开关并使用哈希图来关联计数值和要绘制的位图。数组在这里甚至会更快，也许更合适。

于 2013-10-28T06:25:45.433 回答
2

您应该从您的方法中删除所有BitmapFactory.decodeResource()调用draw()。只解码一次位图并保持对它的引用。然后只需调用canvas.drawBitmap()您的draw()方法。

于 2013-10-28T06:24:22.000 回答