CODEWITHSUNDEEP
pythonhtmlpython-3.xweb-scrapingbeautifulsoup
PURWU
PURWU

Reputation: 427

BeautifulSoup to get image name from P class picture tag in Python

The HTML looks

<p class="rating item-rating">
<picture>
<source srcset="/assets/img/ratings/rating-4_5.svg" type="image/svg+xml"/>
<img src="/assets/img/ratings/rating-4_5.png"/>
</picture>
<span>
260
</span>
</p>

And I would like to get 

/assets/img/ratings/rating-4_5.png

How should I improve the following code?

img = soup.findAll('p',attrs={'class':'rating item-rating'})

for i in img:
    print(i.picture)

Upvotes: 0

Views: 1134

Answers (2)

Omer Tekbiyik
Omer Tekbiyik

Reputation: 4744

You can get src value in img tag easily like :

   import requests

from bs4 import BeautifulSoup
r = """<p class="rating item-rating">
<picture>
<source srcset="/assets/img/ratings/rating-4_5.svg" type="image/svg+xml"/>
<img src="/assets/img/ratings/rating-4_5.png"/>
</picture>
<span>
260
</span>
</p>"""
source = BeautifulSoup(r,'html')

img = source.findAll('p',attrs={'class':'rating item-rating'})

for parsing in img:
    print(parsing.img['src'])

Upvotes: 1

user5386938
user5386938

Reputation:

You need to get to the img tag as that seems to hold the information you want in the src attribute.

from bs4 import BeautifulSoup

s = '''<p class="rating item-rating">
<picture>
<source srcset="/assets/img/ratings/rating-4_5.svg" type="image/svg+xml"/>
<img src="/assets/img/ratings/rating-4_5.png"/>
</picture>
<span>
260
</span>
</p>'''

soup = BeautifulSoup(s, 'html.parser')
for p in soup.select('p.rating'):
    print(p.picture.img['src'])

Upvotes: 2

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