打开PopupWindow，让外侧仍然可触摸 [英] Open a PopupWindow and let the outsides still touchable

问题描述

如何打开Android上PopupWindow，让所有其他部件可触及不驳回PopupWindow？

这是它是如何创建的：

 公共类DynamicPopup {
    私人最终PopupWindow窗口;
    私人最终RectF RECT;
    私人最终查看父母;
    私人最终RichPageView图。    公共DynamicPopup（上下文的背景下，RichPage页，RectF rectF，查看父）{
        this.parent =父母;
        RECT = rectF;        窗口=新PopupWindow（上下文）;        window.setBackgroundDrawable（新BitmapDrawable（））;
        window.setWidth（（int）的rect.width（））;
        window.setHeight（（int）的rect.height（））;
        window.setTouchable（真）;
        window.setFocusable（真）;
        window.setOutsideTouchable（真）;        鉴于=新RichPageView（背景下，页面，假）;
        window.setContentView（视图）;        view.setOnCloseListener（新侦听器（）{
            @覆盖
            公共无效的OnAction（）{
                window.dismiss（）;
            }
        }）;
    }    公共无效显示（）{
        window.showAtLocation（父母，Gravity.NO_GRAVITY，（INT）rect.left，（INT）rect.top）;
    }
}
 

解决方案

根据的javadoc

  

控制是否弹出将触摸事件的其窗口之外被告知。这仅是有道理的弹出窗口是可触摸的，但不可作为焦点

所以你的行

  window.setFocusable（真）;
 

则此方法 setOutsideTouchable（）什么都不做。

How to open a PopupWindow on Android and let all the others components touchable without dismiss the PopupWindow?

This is how it's created:

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);
    }
}

解决方案

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

So your line

 window.setFocusable(true);

causes the method setOutsideTouchable() to do nothing.

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