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I have link "https://uchebnik.mos.ru/app_player/204597". I need to get url with "https://learningapps.org/show.php?id=pxnu062wj20&fullscreen=1" or "pxnu062wj20". I have found this data in network: enter image description here

And in "sources" tab: enter image description here

It is desirable that this be done without using a webdriver. Also, i can use requests, requests_html, urllib3.

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  • That is great. However, what is your current progress on this? Could you post any code that you tried and it fails for some reason? Commented Dec 24, 2021 at 18:35
  • I don't have any code that relates to this. I tried to receive html page content, send GET requests, but I don't know how to get what I need. Commented Dec 24, 2021 at 18:38
  • You can try this yourself https://docs.python-requests.org/en/latest/ Commented Dec 24, 2021 at 18:43
  • How can the link to the documentation of this module help me? I've already tried looking for answers but couldn't find anything. Commented Dec 24, 2021 at 18:50
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    We help you code, we don't do the coding for you. So please show your best attempt so we can help you out. Make sure to include a minimal, reproducible example, now I get an error message when opening the site ("нет доступа к приложению") Commented Dec 24, 2021 at 18:56
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