2

前几天我试图从 DDG 抓取搜索结果，但我不断收到响应 418。我怎样才能让它响应 200 或从中获取结果？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)```
4

1 回答 1

1

您可以使用本地化的 DDG 的仅 HTML 版本 ( https://html.duckduckgo.com/html/) 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))

印刷：

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于 2020-07-01T11:14:39.827 回答