BeautifulSoup 从 Python 中的 P 类图片标签中获取图像名称答案

【问题标题】：BeautifulSoup to get image name from P class picture tag in PythonBeautifulSoup 从 Python 中的 P 类图片标签中获取图像名称
【发布时间】：2020-06-10 13:57:28
【问题描述】：

HTML 外观

<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>

我想得到

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

我应该如何改进以下代码？

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

for i in img:
    print(i.picture)

【问题讨论】：

标签： python html python-3.x web-scraping beautifulsoup

【解决方案1】：

您需要访问img 标记，因为它似乎在src 属性中保存了您想要的信息。

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'])

【讨论】：

【解决方案2】：

您可以在img 标签中轻松获取src 值，例如：

   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'])

【讨论】：