[英]How to close a fancybox element in Google Chrome developer tools?
I would like to close a popup content in some web pages via Chrome developer console. 我想通过Chrome开发者控制台关闭某些网页中的弹出式内容。
For example: When I open http://www.cw.com.tw for the first time, it will show a fancybox-content
element. 例如:当我第一次打开http://www.cw.com.tw时,它将显示
fancybox-content
元素。
So I will have to click the "close button" of the advertisement in order to read to article. 因此,我必须单击广告的“关闭按钮”才能阅读文章。
This website is trustable, but for some websites, if I don't want to click that element for some security concerns, how can I close that popup content and remove the grey layer without click the fancybox-close
button . 该网站是可信任的,但是对于某些网站,如果出于安全考虑我不想单击该元素, 那么如何在不单击
fancybox-close
按钮的情况下关闭该弹出内容并删除灰色层 。
In another word, how can I know which function it triggers when I click that fancybox-close
button? 换句话说,当我单击
fancybox-close
按钮时,如何知道它触发了哪个功能?
I tried inspected the element, and deleted the fancybox-content
node directly. 我尝试检查了该元素,然后直接删除了
fancybox-content
节点。 But it didn't help very much. 但这并没有太大帮助。
Any help would be appreciated, thanks in advance. 任何帮助将不胜感激,在此先感谢。
You should have read the document first though, 不过您应该先阅读该文档,
$.fancybox.close()
should do the trick. 应该可以。
FYI, the API documentation here. 仅供参考,此处是API文档。
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