[FIXED] Python: scraping google results for websites' main URL and title

Issue

I am trying to scrape a given number results from google search, but I so far I came across two problems: one is that I don't know how to join the URLs and the titles inside the same loop, so they can be shown together in the format:

(Title)
(Website URL)
(---------)
(Title)
(Website URL)
(---------)

I somehow managed to achieve this format, but the loop is going on several times, instead of just showing the top 10 results. I believe it's something to do with how I structured the loops to work together, but I don't know how to avoid this.

The other problem is that I want to obtain both main URL and title of each website within search results, but while I managed to get the right titles, I seem to be getting many links coming from the same website, instead of only the main URL. For instance, if I search for "data science", the second or third title shown is from Coursera, while the link is from wikipedia. I only want the main URL so the title matches the website URL, how do I get it?

Any input will be greatly appreciated

import requests
from bs4 import BeautifulSoup
import re

query = "data science"
search = query.replace(' ', '+')
results = 10
url = (f"https://www.google.com/search?q={search}&num={results}")

requests_results = requests.get(url)
soup_link = BeautifulSoup(requests_results.content, "html.parser")
soup_title = BeautifulSoup(requests_results.text,"html.parser")
links = soup_link.find_all("a")
heading_object=soup_title.find_all( 'h3' )

for link in links:
  for info in heading_object:
    get_title = info.getText()
    link_href = link.get('href')
    if "url?q=" in link_href and not "webcache" in link_href:
      print(get_title)
      print(link.get('href').split("?q=")[1].split("&sa=U")[0])
      print("------")

Solution

The length of your links doesn't seem to match your heading_object list. I think it's best if you filter it further than just "a".

Editing your solution, you can loop through links like this:

import requests
from bs4 import BeautifulSoup
import re

query = "data science"
search = query.replace(' ', '+')
results = 10
url = (f"https://www.google.com/search?q={search}&num={results}")

requests_results = requests.get(url)
soup_link = BeautifulSoup(requests_results.content, "html.parser")
links = soup_link.find_all("a")

for link in links:
    link_href = link.get('href')
    if "url?q=" in link_href and not "webcache" in link_href:
      title = link.find_all('h3')
      if len(title) > 0:
          print(link.get('href').split("?q=")[1].split("&sa=U")[0])
          print(title[0].getText())
          print("------")

Instead of keeping 2 lists for headers and links, we can get the header directly from the link. We do that by by doing another find_all('h3') inside the link object. Since there are links that match url?q= format but are not part of the actual results you want to display, like the expanding accordion for related searches etc, we need to filter those out too. We can do that by checking if they have an "h3" header that's why we have len(title) > 0.

Answered By - Jerven Clark