如何排除某些我不想要的beautifulsoup结果？

Question

我在尝试排除Beauty soup程序中给出的结果时遇到问题这是我的代码：

from bs4 import BeautifulSoup
import requests

URL = 'https://en.wikipedia.org/wiki/List_of_Wikipedia_mobile_applications'
page = requests.get(URL)

soup = BeautifulSoup(page.content, 'html.parser')

for link in soup.find_all('a'):
    print(link.get('href'))

我不想得到以“#”开头的结果，例如：#cite#u ref-18

我尝试使用for循环，但收到以下错误消息：KeyError: 0

Answer 1

您可以使用str.startswith()方法：

from bs4 import BeautifulSoup
import requests

URL = 'https://en.wikipedia.org/wiki/List_of_Wikipedia_mobile_applications'
page = requests.get(URL)

soup = BeautifulSoup(page.content, 'html.parser')

for tag in soup.find_all('a'):
    link = tag.get('href')
    if not str(link).startswith('#'):
        print(link)

Answer 2

您可以使用CSS选择器a[href]:not([href^="#"])。这将选择具有href=属性的所有<a>标记，但不选择以#字符开头的标记：

import requests
from bs4 import BeautifulSoup

URL = 'https://en.wikipedia.org/wiki/List_of_Wikipedia_mobile_applications'
page = requests.get(URL)

soup = BeautifulSoup(page.content, 'html.parser')

for link in soup.select('a[href]:not([href^="#"])'):
    print(link['href'])