为 this 问题提供部分答案,我来了bs4.element.Tag
是一堆嵌套的字典和列表(s
,下面).
Working on a partial answer to this question, I came across a bs4.element.Tag
that is a mess of nested dicts and lists (s
, below).
有没有办法使用 re.find_all
返回包含在 s
中的 url 列表?有关此标签结构的其他评论也很有帮助.
Is there a way to return a list of urls contained in s
without using re.find_all
? Other comments regarding the structure of this tag are helpful too.
from bs4 import BeautifulSoup
import requests
link = 'https://stackoverflow.com/jobs?med=site-ui&ref=jobs-tab&sort=p'
r = requests.get(link)
soup = BeautifulSoup(r.text, 'html.parser')
s = soup.find('script', type='application/ld+json')
## the first bit of s:
# s
# Out[116]:
# <script type="application/ld+json">
# {"@context":"http://schema.org","@type":"ItemList","numberOfItems":50,
我尝试过的:
s
上随机浏览带有 tab 补全的方法.s
.我的问题是 s
只有 1 个属性(type
)而且似乎没有任何子标签.
My problem is that s
only has 1 attribute (type
) and doesn't seem to have any child tags.
可以使用s.text
来获取脚本的内容.它是 JSON,因此您可以使用 json.loads
对其进行解析.从那里,它是简单的字典访问:
You can use s.text
to get the content of the script. It's JSON, so you can then just parse it with json.loads
. From there, it's simple dictionary access:
import json
from bs4 import BeautifulSoup
import requests
link = 'https://stackoverflow.com/jobs?med=site-ui&ref=jobs-tab&sort=p'
r = requests.get(link)
soup = BeautifulSoup(r.text, 'html.parser')
s = soup.find('script', type='application/ld+json')
urls = [el['url'] for el in json.loads(s.text)['itemListElement']]
print(urls)
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