BS4 Python 获取 href url

Question

我堆叠了 bs4 脚本，我需要获取 href 链接或元内容，我该怎么做？基本上我需要得到这个：<meta itemprop="image" content="https://resources.reed.co.uk/profileimages/logos/thumbs/Logo_71709.png?v=20200828172950">或者<img src="https://resources.reed.co.uk/profileimages/logos/thumbs/Logo_71709.png?v=20200828172950" alt="Posted by Publica Group " width="120" height="50" class=" b-loaded" style="display: inline;">我尝试这样做：            logoscrap = soup.find('meta', attrs={'itemprop': 'image'})和            logoscrap = soup.find('img', class_="b-loaded").attrs['src']但我的代码不起作用...

Answer 1

soup.find 返回 dict 对象，您可以直接从 dict 访问属性

img = soup.find('meta', attrs={'itemprop': 'image'})

logoscrap = img['content']

#output:

https://resources.reed.co.uk/profileimages/logos/thumbs/Logo_71709.png?v=20200828172950

或者

img = soup.find('img', class_="b-loaded")

logoscrap  = img['src']

#output:

https://resources.reed.co.uk/profileimages/logos/thumbs/Logo_71709.png?v=20200828172950