Question

我正在尝试使用LayerDrawable作为源来扩展ImageView。每一层都包含一个StateListDrawable，其中一个是ShapeDrawable，另一个是从资源加载的Drawable。

在用户按下时，ShapeDrawable从青色渐变为白色，并且图标可从其原始颜色（白色）绘制为可绘制的颜色过滤器，将其转换为相同的青色。

图标始终保持原始颜色相同，但ShapeDrawable渐变为白色。我已经从图标中删除了白色状态，并且出现了青色，因此我假设它在StateListDrawable的某个地方出现问题。我也删除了进入和退出淡入淡出持续时间而没有任何成功。

代码如下

public class DrawerItemIcon extends ImageView {

    private static final String ITEM_ENABLED_COLOR = "#1ABC9C";

    private int mIconResource;
    private StateListDrawable mCircleDrawable;
    private StateListDrawable mIconDrawable;

    public DrawerItemIcon(Context context, int iconResource) {
        super(context);
        mIconResource = iconResource;
        init();
    }

    private void init() {
        setDuplicateParentStateEnabled(true);
        setImageDrawable(getLayers());
    }

    private LayerDrawable getLayers() {
        Drawable[] layers = new Drawable[2];
        layers[1] = getCircleDrawable();
        layers[0] = getIconDrawable();
        return new LayerDrawable(layers);
    }

    private Drawable getCircleDrawable() {
        mCircleDrawable = new StateListDrawable();
        mCircleDrawable.addState(new int[]{android.R.attr.state_pressed},getPressedShapeCircleDrawable());
        mCircleDrawable.addState(new int[]{android.R.attr.state_enabled}, getEnabledShapeCircleDrawable());
        mCircleDrawable.setEnterFadeDuration(500);
        mCircleDrawable.setExitFadeDuration(500);
        return mCircleDrawable;
    }

    private ShapeDrawable getPressedShapeCircleDrawable() {
        ShapeDrawable bgDrawable = new ShapeDrawable();
        bgDrawable.setShape(new OvalShape());
        bgDrawable.getPaint().setStyle(Paint.Style.STROKE);
        bgDrawable.getPaint().setStrokeWidth(4);
        bgDrawable.getPaint().setColor(Color.WHITE);
        return bgDrawable;
    }

    private ShapeDrawable getEnabledShapeCircleDrawable() {
        ShapeDrawable bgDrawable = new ShapeDrawable();
        bgDrawable.setShape(new OvalShape());
        bgDrawable.getPaint().setStyle(Paint.Style.STROKE);
        bgDrawable.getPaint().setStrokeWidth(4);
        bgDrawable.getPaint().setColor(Color.parseColor(ITEM_ENABLED_COLOR));
        return bgDrawable;
    }

    public Drawable getIconDrawable() {
        mIconDrawable = new StateListDrawable();
        mIconDrawable.addState(new int[]{android.R.attr.state_pressed}, getPressedIconDrawable());
        mIconDrawable.addState(new int[]{android.R.attr.state_enabled}, getEnabledIconDrawable());
        mIconDrawable.setEnterFadeDuration(500);
        mIconDrawable.setExitFadeDuration(500);
        return mIconDrawable;
    }

    private Drawable getPressedIconDrawable() {
        return getResources().getDrawable(mIconResource);
    }

    private Drawable getEnabledIconDrawable() {
        Drawable selectedDrawable = getResources().getDrawable(mIconResource);
        selectedDrawable.setColorFilter(
        new PorterDuffColorFilter(Color.parseColor(ITEM_ENABLED_COLOR), PorterDuff.Mode.MULTIPLY));
        return selectedDrawable;
    }
}

Answer 1

这是因为那个可绘制的常态。解决此问题的方法是在.mutate()方法中添加getEnabledIconDrawable()来电。

... getResources().getDrawable(mIconResource).mutate();

有关常量状态的更多信息，请访问：http://www.curious-creature.com/2009/05/02/drawable-mutations/

在其他地方回答，但会放在这里供将来参考。

带有LayerDrawable的ImageView包含两个StateListDrawable，而不是在一个图层上绘制按下状态

1 个答案: