android - 开个弹窗让外人照样可以摸

Question

如何在 Android 上打开 PopupWindow 并在不关闭 PopupWindow 的情况下让所有其他组件可触摸？

它是这样创建的:

public class DynamicPopup {
    private final PopupWindow window;
    private final RectF rect;
    private final View parent;
    private final RichPageView view;

    public DynamicPopup(Context context, RichPage page, RectF rectF, View parent) {
        this.parent = parent;
        rect = rectF;

        window = new PopupWindow(context);

        window.setBackgroundDrawable(new BitmapDrawable());
        window.setWidth((int) rect.width());
        window.setHeight((int) rect.height());
        window.setTouchable(true);
        window.setFocusable(true);
        window.setOutsideTouchable(true);

        view = new RichPageView(context, page, false);
        window.setContentView(view);

        view.setOnCloseListener(new Listener(){
            @Override
            public void onAction() {
                window.dismiss();
            }
        });


    }

    public void show() {
        window.showAtLocation(parent, Gravity.NO_GRAVITY, (int) rect.left, (int) rect.top);
    }
}

Answer 1

就像 ernazm 说的那样

As per javadocs

Controls whether the pop-up will be informed of touch events outside of its window. This only makes sense for pop-ups that are touchable but not focusable

正在努力

window.setTouchable(true);
window.setFocusable(false);
window.setOutsideTouchable(false);

当窗口touchalbe为true，focusable为false时，setOutsideTouchable()有效，如果setOutsideTouchable(true)，弹出窗口外的触摸将被关闭，否则弹出窗口外仍然可以触摸而不关闭。