当弹出窗口处于活动状态时，如何与离子中的项目进行交互？
Ionic版本：1.x
Ionic Version: 1.x
平台：全部
我有一个离子应用程序使用了离子弹出。但是，当弹出窗口处于活动状态（打开）时，我无法再与后台的内容进行交互，例如向下滚动我的项目列表。我已经尝试寻找这个问题的现有答案，但没有看到问题。
I have an ionic app which uses an ionic popover. However, when the popover is active (open), I can no longer interact with things in the background, for example scroll down my list of items. I have tried looking for existing answers for this question, but did not see the question being asked.
popover本身不必绑定到任何活动元素页面，因为它当前有位置：固定并且总是出现在同一位置。
The popover itself does not have to be bound to any active element on the page, since it currently has position:fixed and will always appear in the same position.
所以基本上问题是，有没有办法防止弹出窗口阻止我与背景中的所有内容进行交互？
So basically the question is, is there a way to prevent the popover from preventing me interacting with everything in the background?
我找到了答案，但不确定是否有更好的方法可以做这可能是一个选项，可以打开和关闭与背景的交互，但现在答案似乎如下：
I have found an answer, however, not sure if there is a better way to do this, perhaps an option that turns interacting with the background on and off, but for now the answer seems to be as follows:
- 确保.popover-backdrop类与实际固定的popover元素具有相同的大小和位置（在我的情况下是高度和上边距）。默认值为height 100％。
- 确保对.popover-open类（应用于正文）禁用
pointer-events：none; - 请记住，这将改变行为，如果您在弹出窗口外单击它将默认关闭它。所以在离子方面，我必须确保在我离开视图时关闭弹出窗口的事件如果处于活动状态。