BS4 Python得到一个href url

Question

我在 bs4 脚本上堆叠，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

<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;">

我试过这样做：

            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，你可以直接从 dict 获取属性

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

img = soup.find('img', class_="b-loaded")
logoscrap  = img['src']
#output:
https://resources.reed.co.uk/profileimages/logos/thumbs/Logo_71709.png?v=20200828172950