如何通过 Beautiful Soup 抓取 href 中的文本?

How to scrape text in a href by Beautiful Soup?

我有一个格式为 <a href="javascript:ShowImg('../UploadFile/Images/c/1/B_27902.jpg');"> 的 href,我想用 '../UploadFile/Images/c/1/B_27902.jpg' 获得 url。我用了很笨的方法来获取:(我想知道有没有更简单的获取方法

url = '<a href="javascript:ShowImg('../UploadFile/Images/c/1/B_27902.jpg');">'
html = url.get('href')
html = html.replace('javascript:ShowImg(', '').replace(');', '')

原标签如下:

<a href="javascript:ShowImg('../UploadFile/Images/c/1/B_27902.jpg');">
<img height="110" onerror="this.src='../UploadFile/Images/no_pic_big.jpg';"
src="../UploadFile/Images/c/1/S_27902.jpg" width="170"/>
</a>

BeautifulSoup 可以在搜索元素时将 compiled regular expression pattern 应用于属性值。然后您可以使用相同的模式来提取所需的部分:

import re
from bs4 import BeautifulSoup

data = """
<a href="javascript:ShowImg('../UploadFile/Images/c/1/B_27902.jpg');">
<img height="110" onerror="this.src='../UploadFile/Images/no_pic_big.jpg';"
src="../UploadFile/Images/c/1/S_27902.jpg" width="170"/>
</a>
"""

soup = BeautifulSoup(data, "html.parser")
pattern = re.compile(r"javascript:ShowImg\('(.*?)'\);")

href = soup.find('a', href=pattern)["href"]
link = pattern.search(href).group(1)
print(link)  # prints ../UploadFile/Images/c/1/B_27902.jpg