获取响应 200 而不是 <418 我是茶壶>,使用 DDG



前几天我试图从DDG中抓取搜索结果,但我不断收到回复418。如何使其响应 200 或从中获得结果?这是我的代码。

import requests
from bs4 import BeautifulSoup
import urllib

while True:

query = input("Enter Search Text: ")
a = query.replace(' ', '+')
url = 'https://duckduckgo.com/?q=random' +a

headers = {"User-Agent": "Mozilla/5.0 (Linux; Android 6.0.1; SHIELD Tablet K1 Build/MRA58K; wv) AppleWebKit/537.36 (KHTML, like Gecko) Version/4.0 Chrome/55.0.2883.91 Safari/537.36"}

r = requests.get(url)
print(r)
soup = BeautifulSoup(r.content,'lxml')

tags = soup.findAll('h2', class_="result__title")
print(tags)```

您可以使用(https://html.duckduckgo.com/html/( 本地化的 DDG 的纯 HTML 版本来获取结果。

例如:

import requests
from bs4 import BeautifulSoup

url = 'https://html.duckduckgo.com/html/'
params = {'q': 'python'}
headers = {'User-Agent': 'Mozilla/5.0 (X11; Ubuntu; Linux x86_64; rv:77.0) Gecko/20100101 Firefox/77.0'}
soup = BeautifulSoup(requests.get(url, params=params, headers=headers).content, 'html.parser')
for t in soup.select('h2.result__title'):
print(t.get_text(strip=True))

指纹:

LearnPythonOnline - Start Today & Change Your LifeAdViewing ads is privacy protected by DuckDuckGo. Ad clicks are managed by Microsoft's ad network (more info).
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