DuckDuckGo结果抓取

Question

我在运行我的代码时遇到了问题，并在StackOverflow上找到了一个完美的解决方案。但是，当我进行必要的更改并运行它时，我得不到任何输出。

代码：

from bs4 import BeautifulSoup
import urllib.parse
import requests

r = requests.get('https://duckduckgo.com/html/?q=test')
soup = BeautifulSoup(r.text, 'html.parser')
results = soup.find_all('a', attrs={'class':'result__url'}, href=True)

for link in results:
    url = link['href']
    o = urllib.parse.urlparse(url)
    d = urllib.parse.parse_qs(o.query)
    print(d['uddg'][0])

路径组件的uddg ()“获取查询字符串并将其传递给urlparse进一步处理，然后可以使用urlparse名称提取链接。”这应该是最初的几个结果：

http://www.speedtest.net/
https://www.merriam-webster.com/dictionary/test
https://en.wikipedia.org/wiki/Test
https://www.thefreedictionary.com/test
https://www.dictionary.com/browse/test

我没有得到任何输出。输出：

In [13]: runfile('C:/Users/Spurs/.spyder-py3/temp.py', wdir='C:/Users/Spurs/.spyder-py3')
In [14]:

Answer 1

您将获得一个403，因此您没有任何结果。要解决此问题，请添加headers。

下面是操作步骤：

import requests
from bs4 import BeautifulSoup

headers = {
    "User-Agent": "Mozilla/5.0 (Macintosh; Intel Mac OS X 10.15; rv:84.0) Gecko/20100101 Firefox/84.0",
}

page = requests.get('https://duckduckgo.com/html/?q=test', headers=headers).text
soup = BeautifulSoup(page, 'html.parser').find_all("a", class_="result__url", href=True)

for link in soup:
    print(link['href'])

输出：

https://www.merriam-webster.com/dictionary/test
https://www.speedtest.net/
https://www.dictionary.com/browse/test
https://www.thefreedictionary.com/test
https://www.thesaurus.com/browse/test
https://en.wikipedia.org/wiki/Test
https://www.tests.com/
http://speedtest.xfinity.com/
https://fast.com/
https://www.spectrum.com/internet/speed-test
https://projectstream.google.com/speedtest
https://dictionary.cambridge.org/dictionary/english/test
http://www.act.org/content/act/en/products-and-services/the-act.html
...