Как спарсить атрибут src через beautiful soup?

Question

[cognito33]

Нужно спарсить ./3904450_files/5211588560.jpg

<img class="d-block" src="./3904450_files/5211588560.jpg" itemprop="image" alt="Корпус xiaomi redmi 4x"></div>

Comments

[Рамис]

>>> from bs4 import BeautifulSoup
>>> html = """<img class="d-block" src="./3904450_files/5211588560.jpg" itemprop="image" alt="Корпус xiaomi redmi 4x"></div>"""
>>> bs = BeautifulSoup(html, 'html.parser')
>>> img = bs.find('img')
>>> img['src']
'./3904450_files/5211588560.jpg'