How to get the link from content='link'?

H

houne2021-12-21 02:18:48

Python

houne, 2021-12-21 02:18:48

Let's say there is an element:

<meta property="og:audio" content="https://freesound.org/data/previews/611/611605_1648170-lq.mp3" />

How do I get into a variable

https://freesound.org/data/previews/611/611605_1648170-lq.mp3

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1 answer(s)

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Sergey Karbivnichy, 2021-12-21
@hottabxp

from bs4 import BeautifulSoup

html = '<meta property="og:audio" content="https://freesound.org/data/previews/611/611605_1648170-lq.mp3" />'

soup = BeautifulSoup(html,"lxml")
mp3_url = soup.find('meta',property='og:audio').get('content')
print(mp3_url)

>>> https://freesound.org/data/previews/611/611605_1648170-lq.mp3