python - 如何使用 Beautifulsoup 访问前五个谷歌结果链接

Question

我想访问来自 Google 的前五个(或任何指定数量的)结果链接。通过研究，我发现并修改了如下代码。

import requests
from bs4 import BeautifulSoup
import re    
search = raw_input("Search:")
page = requests.get("https://www.google.com/search?q=" + search)
soup = BeautifulSoup(page.content, "lxml")
links = soup.find("a")
print links.get('href')

这将返回页面上的第一个链接，似乎每次都是 Google 图片选项卡。

这不完全是我想要的。对于初学者，我不想要任何谷歌网站的链接，只想要结果。另外，我想要前三个或五个或任何指定数量的结果。

我如何使用 python 来执行此操作？

提前致谢!

Answer 1

您可以使用:

import requests
from bs4 import BeautifulSoup
import re
search = input("Search:")
results = 100 # valid options 10, 20, 30, 40, 50, and 100
page = requests.get(f"https://www.google.com/search?q={search}&num={results}")
soup = BeautifulSoup(page.content, "html5lib")
links = soup.findAll("a")
for link in links :
    link_href = link.get('href')
    if "url?q=" in link_href and not "webcache" in link_href:
        print (link.get('href').split("?q=")[1].split("&sa=U")[0])

Google Search Demo

对于 duckduckgo.com 使用:

import requests
from bs4 import BeautifulSoup
import re
search = input("Search:")
h = {"Host":"duckduckgo.com", "Origin": "https://duckduckgo.com", "User-Agent": "Mozilla/5.0 (Windows NT 10.0; Win64; x64; rv:74.0) Gecko/20100101 Firefox/74.0"}
d = {"q":search}
page = requests.post(f"https://duckduckgo.com/html/", data=d, headers=h)
soup = BeautifulSoup(page.content, "html5lib")
links = soup.findAll("a", {"class": "result__a"})
for link in links :
    link_href = link.get('href')
    if not "https://duckduckgo.com" in link_href:
        print(link_href)