Day: March 31, 2014

Getting Google Search results with Scrapy

Google do not allow easy scraping of their search results. As Google, they are smart to detect bots and prevent them from scraping the results automatically. The following will attempt to scrape search results based on python Scrapy. The full script for this project is not completed and will be included in subsequent posts.

Scrapy make use of the starting url for google search. Example is a format used by google to search a particular keyword.

More details on the url construction can be found in the following link.

With the URL constructed, the web link results related to the search can be pulled from stand-alone scrapy spider. The xpath specified in the scrapy spider is the html tags that the the link results resides in.The xpath expression is as below:

sel = Selector(response)
## extract a list of website link related to the search
google_search_links_list = sel.xpath('//h3/a/@href').extract()

Only Link results are extracted based on current plan . As the format of google search is consistently changing, it is more difficult to retrieve other information. The plan is to extract the links and then access the individual links using scrapy and retrieved relevant information. This will be touched on in the subsequent posts.

Example of Scrapy spider used for scraping the google url.
Not actual running code.
import re
import os
import sys
import json

from scrapy.spider import Spider
from scrapy.selector import Selector

class GoogleSearch(Spider):

 #set the search result here
 name = 'Google search'
 allowed_domains = ['']
 start_urls = ['Insert the google url here']

 def parse(self, response):

 sel = Selector(response)
 google_search_links_list = sel.xpath('//h3/a/@href').extract()
 google_search_links_list = ['q=(.*)&sa',n).group(1) for n in google_search_links_list]

## Dump the output to json file
 with open(output_j_fname, "w") as outfile:
 json.dump({'output_url':google_search_links_list}, outfile, indent=4)