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?
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