Selenium Python: How to get css without targetting a specific class/id/tag

Question

I'm working on a scraper project and one of the goals is to get every image link from HTML & CSS of a website. I was using BeautifulSoup & TinyCSS to do that but now I'd like to switch everything on Selenium as I can load the JS.

I can't find in the doc a way to target some CSS parameters without having to know the tag/id/class. I can get the images from the HTML easily but I need to target every "background-image" parameter from the CSS in order to get the URL from it.

ex: background-image: url("paper.gif");

Is there a way to do it or should I loop into each element and check the corresponding CSS (which would be time-consuming)?

"get every image from HTML & CSS..." What do you mean by get image from CSS? What is your expected result? Do you need to get files of links to images? — Andersson
– Andersson, Commented Sep 28, 2018 at 15:26
Sorry, I just edited the question, I just need the links of every image from a website and not just the ones from the HTML tag "img" — jjyoh
– jjyoh, Commented Sep 28, 2018 at 15:33

José Castro · Accepted Answer · 2018-09-28 16:48:12Z

1

You can grab all the Style tags and parse them, searching what you look.

Also you can download the css file, using the resource URL and parse them.

Also you can create a XPATH/CSS rule for searching nodes that contain the parameter that you're looking for.

answered Sep 28, 2018 at 16:48

José Castro

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1 Comment

jjyoh Over a year ago

Downloading the css file and parse them, I'm already doing it but it takes a bit too much processing time. Could you explain me more about the XPATH/CSS rule, because I tried that and I thought that I would need the class/id/name of the element in order to get the CSS from it. Also could you enlight me about your first option "grab all the Style tages"?

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