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Python 从beautifulsoup中的自定义标记获取url$_Python_Django_Web Scraping_Beautifulsoup - Fatal编程技术网
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Python 从beautifulsoup中的自定义标记获取url$

python django web-scraping

Python 从beautifulsoup中的自定义标记获取url$,python,django,web-scraping,beautifulsoup,Python,Django,Web Scraping,Beautifulsoup,我需要从自定义html标记获取url。 所以我像这样提取img qwe = [] for x in carList: try: carPicture = x.find('img', class_="lazyload", attrs={"data-src":True}) except: carPicture = static("images/human.jpg") qwe.append(carPic

我需要从自定义html标记获取url。 所以我像这样提取img

qwe =   []
for x in carList:
  try:
     carPicture  = x.find('img', class_="lazyload", attrs={"data-src":True})
  except: 
     carPicture   = static("images/human.jpg")
qwe.append(carPicture)
其结果是：

<img class="lazyload" data-src="https://prod.pictures.autoscout24.net/listing-images/58878dbf-083a-4685-af3a-f745a6534426_cdde08fb-999b-488a-86c8-d252bcadaaff.jpg/420x315.jpg"/>
如何仅从数据src标记获取链接？

从，您可以使用以下代码段

carPicture[“数据src”]

输出：

https://prod.pictures.autoscout24.net/listing-images/58878dbf-083a-4685-af3a-f745a6534426_cdde08fb-999b-488a-86c8-d252bcadaaff.jpg/420x315.jpg
这回答了你的问题吗？