Follow First 5 Pages of Google Results Python Mechanize

Question

I'm currently grabbing only the first page of google results for a query, but I want to grab the first 5 pages.

gets a string like: https://encrypted.google.com/search?hl=en&q=site%3Asomedomain.com&start=0

the variable urls gets all 10 results for the first page, but I started adding conditions to check for 10 urls on this first page, if that is true and there are 10 urls, I want it to keep going to the next url e.g. (provided the next url has 10 results as well) using something like follow_link() and urls below :

https://encrypted.google.com/search?hl=en&q=site%3Asomedomain.com&start=10
https://encrypted.google.com/search?hl=en&q=site%3Asomedomain.com&start=20
https://encrypted.google.com/search?hl=en&q=site%3Asomedomain.com&start=30
https://encrypted.google.com/search?hl=en&q=site%3Asomedomain.com&start=40
https://encrypted.google.com/search?hl=en&q=site%3Asomedomain.com&start=50

How do I go about doing this? Could anyone please help me out?

Answer 1

You can use BeautifulSoup to locate element with link to the next page:

from mechanize import Browser
from bs4 import BeautifulSoup

br = Browser()
br.set_handle_robots(False)
br.addheaders = [('User-agent', 'Mozilla/5.0 (Windows NT 6.2;\
                    WOW64) AppleWebKit/537.11 (KHTML, like Gecko)\
                    Chrome/23.0.1271.97 Safari/537.11')]

url = "https://encrypted.google.com/search?hl=en&q=site%3Asomedomain.com&start=0"

r = br.open(url)

soup = BeautifulSoup(r)

nextpage = soup.find("a", {"id": "pnnext"})
print nextpage['href']

Output:

/search?q=site:somedomain.com&hl=en&ei=NJ4HUo2yM-TK4ATJlYGICQ&start=10&sa=N

So now you have the link to the next page. If element wasn't found then it's the last page