如何用BeautifulSoup提取含空格和换行的class对应的href属性

方法1：用class_参数传入类名列表 #

BeautifulSoup会自动忽略class属性里的换行、多空格，直接把class的所有拆分项作为列表传入class_参数即可定位标签：

from bs4 import BeautifulSoup

html = '''<a class="this is part1 
      this is part2" target="_self" href="https://www.website.com/">'''
soup = BeautifulSoup(html, 'html.parser')

target_link = soup.find('a', class_=['this', 'is', 'part1', 'this', 'is', 'part2'])
if target_link:
    print(target_link['href'])  # 输出：https://www.website.com/

方法2：使用CSS选择器 #

用.连接所有类名组成CSS选择器，不管原class里的空白符格式：

from bs4 import BeautifulSoup

html = '''<a class="this is part1 
      this is part2" target="_self" href="https://www.website.com/">'''
soup = BeautifulSoup(html, 'html.parser')

target_link = soup.select_one('a.this.is.part1.this.is.part2')
if target_link:
    print(target_link['href'])  # 输出：https://www.website.com/

内容的提问来源于stack exchange，提问作者jpy