How to access top five Google result links using Beautifulsoup

Question

I want to access the top five(or any specified number) of links of results from Google. Through research, I found and modified the following code.

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')

This returns the first link on the page, which seems to be the Google images tab every time.

This is not completely what I want. For starters, I don't want the links of any google sites, just the results. Also, I want the first three or five or any specified number of results.

How can I use python to do this?

Thanks ahead of time!

Answer 1

You can use:

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

For duckduckgo.com use:

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)