【发布时间】:2020-10-21 18:43:42
【问题描述】:
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)```
【问题讨论】:
标签: python web-scraping beautifulsoup python-requests duckduckgo