hellpanderrr hellpanderrr - 1 year ago 203
Python Question

Scrapy callback after redirect

I have a very basic scrapy spider, which grabs urls from the file and then downloads them. The only problem is that some of them got redirected to a slightly modified url within same domain. I want to get them in my callback function using response.meta, and it works on a normal urls, but then url is redirected callback doesn't seem to get called. How can I fix it?
Here's my code.

from scrapy.contrib.spiders import CrawlSpider
from scrapy import log
from scrapy import Request
class DmozSpider(CrawlSpider):
name = "dmoz"
handle_httpstatus_list = [302]
allowed_domains = ["http://www.exmaple.net/"])
f = open("C:\\python27\\1a.csv",'r')
url = 'http://www.exmaple.net/Query?indx='
start_urls = [url+row for row in f.readlines()]
def parse(self, response):
print response.meta.get('redirect_urls', [response.url])
print response.status
print (response.headers.get('Location'))

I've also tried something like that:

def parse(self, response):
return Request(response.url, meta={'dont_redirect': True, 'handle_httpstatus_list': [302]}, callback=self.parse_my_url)
def parse_my_url(self, response):
print response.status
print (response.headers.get('Location'))

And it doesn't work either.

Answer Source

By default scrapy requests are redirected, although if you don't want to redirect you can do like this, use start_requests method and add flags in request meta.

    def start_requests(self):
        requests =[(Request(self.url+u, meta={'handle_httpstatus_list': [302],
                               'dont_redirect': True},         
                    callback=self.parse)) for u in self.start_urls]
        return requests
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download