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