Android的弹出窗口时，驳斥外界点击 [英] Android Popup Window dismisses when clicked outside

问题描述

我希望得到的答案，我的问题，我目前所面对的。

I was hoping to get an answer to my problem I have at the moment.

我有延伸弹出窗口的类。它工作正常，但我不想解雇，当我点击窗口外的窗口。

I have a class which extends popup window. It works fine except I don't want the window to dismiss when I click outside of the window.

目前，我有 setOutsideTouchable（假）; 但这只是停止窗外的事件，它仍然驳回弹出

At the moment I have setOutsideTouchable(false); but this just stops events outside the window, it still dismisses the popup.

一个对话框有 setCanceledOnTouchOutside（假），有没有类似的东西，我可以使用？

A dialog has setCanceledOnTouchOutside(false), is there something similar I can use?

感谢

推荐答案

确定这样固定到底。

首先取得该弹出坐落在一个相对布局的主要布局。然后放置在顶部全屏留白的布局设计，我不可见，透明。

First made the main layout which the popup sits on a relative layout. Then placed a full screen blank layout on top which I made invisible and transparent.

然后显示弹出显示时，设置全屏幕面板 setVisibility（View.VISIBLE）可见; 和隐藏弹出时被隐藏与 setVisibility（View.GONE）;

Then show when the popup is shown, set the full screen panel visible with setVisibility(View.VISIBLE); and hide when popup is hidden with setVisibility(View.GONE);

也需要从返回true触摸侦听器的布局（停止触摸事件传递回主布局）：

Also need to return true from an on touch listener for the layout with (To stop touch events passing back to the main layout):

blocker.setOnTouchListener(new OnTouchListener() { 
@Override
public boolean onTouch(View v, MotionEvent event) {
return true;
}
});

和给弹出窗口属性：

setTouchable(true);
setOutsideTouchable(false);

干杯

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