I'm trying to crawl a large site. They have a rate limiting system in place. Is it possible to pause scrapy for 10 minutes when it encounter a 403 page? I know I can set a DOWNLOAD_DELAY but I noticed that I can scrape faster by setting a small DOWNLOAD_DELAY and then pause scrapy for a few minutes when it gets 403. This way the rate limiting gets triggered only once every hour or so.
You can write your own retry middleware and put it to middleware.py
from scrapy.downloadermiddlewares.retry import RetryMiddleware
from scrapy.utils.response import response_status_message
from time import sleep
class SleepRetryMiddleware(RetryMiddleware):
def __init__(self, settings):
RetryMiddleware.__init__(self, settings)
def process_response(self, request, response, spider):
if response.status in [403]:
sleep(120) # few minutes
reason = response_status_message(response.status)
return self._retry(request, reason, spider) or response
return super(SleepRetryMiddleware, self).process_response(request, response, spider)
and don't forget change settings.py
DOWNLOADER_MIDDLEWARES = {
'scrapy.downloadermiddlewares.retry.RetryMiddleware': None,
'your_project.middlewares.SleepRetryMiddleware': 100,
}
Scrapy is a Twisted-based Python framework. So, never use time.sleep or pause.until inside it!
Instead, try using Deferred() from Twisted.
class ScrapySpider(Spider):
name = 'live_function'
def start_requests(self):
yield Request('some url', callback=self.non_stop_function)
def non_stop_function(self, response):
parse_and_pause = Deferred() # changed
parse_and_pause.addCallback(self.second_parse_function) # changed
parse_and_pause.addCallback(pause, seconds=10) # changed
for url in ['url1', 'url2', 'url3', 'more urls']:
yield Request(url, callback=parse_and_pause) # changed
yield Request('some url', callback=self.non_stop_function) # Call itself
def second_parse_function(self, response):
pass
More info here: Scrapy: non-blocking pause
Related
I'm new to scrapy and have been trying to develop a spider that scrapes Tripadvisor's things to do page. Trip advisor paginates results with offset so I made it find the last page num, multiply the number of results per page, and loop over a range with a step of 30. However it returns only a fraction of the results its supposed to, and get_details prints out 7 of the 28 pages scraped. I believe what is happening is url redirection on random pages.
Scrapy logs this 301 redirection on the other pages, and it appears to be redirecting to the first page. I tried disabling redirection but that did not work.
2021-03-28 18:46:38 [scrapy.downloadermiddlewares.redirect] DEBUG: Redirecting (301) to <GET https://www.tripadvisor.com/Attractions-g55229-Activities-a_allAttractions.true-Nashville_Davidson_County_Tennessee.html> from <GET https://www.tripadvisor.com/Attractions-g55229-Activities-a_allAttractions.true-oa90-Nashville_Davidson_County_Tennessee.html>
Here's my code for the spider:
import scrapy
import re
class TripadvisorSpider(scrapy.Spider):
name = "tripadvisor"
start_urls = [
'https://www.tripadvisor.com/Attractions-g55229-Activities-a_allAttractions.true-oa{}-Nashville_Davidson_County_Tennessee.html'
]
def parse(self, response):
num_pages = int(response.css(
'._37Nr884k .DrjyGw-P.IT-ONkaj::text')[-1].get())
for offset in range(0, num_pages * 30, 30):
formatted_url = self.start_urls[0].format(offset)
yield scrapy.Request(formatted_url, callback=self.get_details)
def get_details(self, response):
print('url is ' + response.url)
for listing in response.css('div._19L437XW._1qhi5DVB.CO7bjfl5'):
yield {
'title': listing.css('._392swiRT ._1gpq3zsA._1zP41Z7X::text')[1].get(),
'category': listing.css('._392swiRT ._1fV2VpKV .DrjyGw-P._26S7gyB4._3SccQt-T::text').get(),
'rating': float(re.findall(r"[-+]?\d*\.\d+|\d+", listing.css('svg.zWXXYhVR::attr(title)').get())[0]),
'rating_count': float(listing.css('._392swiRT .DrjyGw-P._26S7gyB4._14_buatE._1dimhEoy::text').get().replace(',', '')),
'url': listing.css('._3W_31Rvp._1nUIPWja._17LAEUXp._2b3s5IMB a::attr(href)').get(),
'main_image': listing.css('._1BR0J4XD').attrib['src']
}
Is there a way to get scrapy working for each page? What is causing this problem exactly?
Found a solution. Discovered I needed to handle the redirection manually and disable Scrapy's default middleware.
Here is the custom middleware I added to middlewares.py
from scrapy.downloadermiddlewares.retry import RetryMiddleware
from scrapy.selector import Selector
from scrapy.utils.response import get_meta_refresh
class CustomRetryMiddleware(RetryMiddleware):
def process_response(self, request, response, spider):
url = response.url
if response.status in [301, 307]:
reason = 'redirect %d' % response.status
return self._retry(request, reason, spider) or response
interval, redirect_url = get_meta_refresh(response)
# handle meta redirect
if redirect_url:
reason = 'meta'
return self._retry(request, reason, spider) or response
hxs = Selector(response)
# test for captcha page
captcha = hxs.xpath(
".//input[contains(#id, 'captchacharacters')]").extract()
if captcha:
reason = 'capcha'
return self._retry(request, reason, spider) or response
return response
It is an updated version from this question's top answer.
Scrapy retry or redirect middleware
I've created a python script using scrapy to scrape some information available in a certain webpage. The problem is the link I'm trying with gets redirected very often. However, when I try few times using requests, I get the desired content.
In case of scrapy, I'm unable to reuse the link because I found it redirecting no matter how many times I try. I can even catch the main url using response.meta.get("redirect_urls")[0] meant to be used resursively within parse method. However, it always gets redirected and as a result callback is not taking place.
This is my current attempt (the link used within the script is just a placeholder):
import scrapy
from scrapy.crawler import CrawlerProcess
class StackoverflowSpider(scrapy.Spider):
handle_httpstatus_list = [301, 302]
name = "stackoverflow"
start_url = 'https://stackoverflow.com/questions/22937618/reference-what-does-this-regex-mean'
def start_requests(self):
yield scrapy.Request(self.start_url,meta={"lead_link":self.start_url},callback=self.parse)
def parse(self,response):
if response.meta.get("lead_link"):
self.lead_link = response.meta.get("lead_link")
elif response.meta.get("redirect_urls"):
self.lead_link = response.meta.get("redirect_urls")[0]
try:
if response.status!=200 :raise
if not response.css("[itemprop='text'] > h2"):raise
answer_title = response.css("[itemprop='text'] > h2::text").get()
print(answer_title)
except Exception:
print(self.lead_link)
yield scrapy.Request(self.lead_link,meta={"lead_link":self.lead_link},dont_filter=True, callback=self.parse)
if __name__ == "__main__":
c = CrawlerProcess({
'USER_AGENT': 'Mozilla/5.0',
})
c.crawl(StackoverflowSpider)
c.start()
Question: How can I force scrapy to make a callback using the url that got redirected?
As far as I understand, you want to scrape a link until it stops redirecting and you finally get http status 200
If yes, then you have to first remove handle_httpstatus_list = [301, 302] from your code
Then create a CustomMiddleware in middlewares.py
class CustomMiddleware(object):
def process_response(self, request, response, spider):
if not response.css("[itemprop='text'] > h2"):
logging.info('Desired text not found so re-scraping' % (request.url))
req = request.copy()
request.dont_filter = True
return req
if response.status in [301, 302]:
original_url = request.meta.get('redirect_urls', [response.url])[0]
logging.info('%s is redirecting to %s, so re-scraping it' % (request._url, request.url))
request._url = original_url
request.dont_filter = True
return request
return response
Then your spider should look like something this
class StackoverflowSpider(scrapy.Spider):
name = "stackoverflow"
start_url = 'https://stackoverflow.com/questions/22937618/reference-what-does-this-regex-mean'
custom_settings = {
'DOWNLOADER_MIDDLEWARES': {
'YOUR_PROJECT_NAME.middlewares.CustomMiddleware': 100,
}
}
def start_requests(self):
yield scrapy.Request(self.start_url,meta={"lead_link":self.start_url},callback=self.parse)
def parse(self,response):
answer_title = response.css("[itemprop='text'] > h2::text").get()
print(answer_title)
If you tell me which site you are scraping then I can help you out, you can email me as well which is on my profile
You may want to see this.
If you need to prevent redirecting it is possible by request meta:
request = scrapy.Request(self.start_url,meta={"lead_link":self.start_url},callback=self.parse)
request.meta['dont_redirect'] = True
yield request
Due to documentation this is a way to stop redirecting.
I've written a script in python's scrapy to make a proxied requests using either of the newly generated proxies by get_proxies() method. I used requests module to fetch the proxies in order to reuse them in the script. However, the problem is the proxy my script chooses to use may not be the good one always so sometimes it doesn't fetch valid response.
How can I let my script keep trying with different proxies until there is a valid response?
My script so far:
import scrapy
import random
import requests
from itertools import cycle
from bs4 import BeautifulSoup
from scrapy.http.request import Request
from scrapy.crawler import CrawlerProcess
class ProxySpider(scrapy.Spider):
name = "sslproxies"
check_url = "https://stackoverflow.com/questions/tagged/web-scraping"
proxy_link = "https://www.sslproxies.org/"
def start_requests(self):
proxylist = self.get_proxies()
random.shuffle(proxylist)
proxy_ip_port = next(cycle(proxylist))
print(proxy_ip_port) #Checking out the proxy address
request = scrapy.Request(self.check_url, callback=self.parse,errback=self.errback_httpbin,dont_filter=True)
request.meta['proxy'] = "http://{}".format(proxy_ip_port)
yield request
def get_proxies(self):
response = requests.get(self.proxy_link)
soup = BeautifulSoup(response.text,"lxml")
proxy = [':'.join([item.select_one("td").text,item.select_one("td:nth-of-type(2)").text]) for item in soup.select("table.table tbody tr") if "yes" in item.text]
return proxy
def parse(self, response):
print(response.meta.get("proxy")) #Compare this to the earlier one whether they both are the same
def errback_httpbin(self, failure):
print("Failure: "+str(failure))
if __name__ == "__main__":
c = CrawlerProcess({
'USER_AGENT': 'Mozilla/5.0',
'DOWNLOAD_TIMEOUT' : 5,
})
c.crawl(ProxySpider)
c.start()
PS My intension is to seek any solution the way I've started here.
As we know http response needs to pass all middlewares in order to reach spider methods.
It means that only requests with valid proxies can proceed to spider callback functions.
In order to use valid proxies we need to check ALL proxies first and after that choose only from valid proxies.
When our previously chosen proxy doesn't work anymore - we mark this proxy as not valid and choose new one from remaining valid proxies in spider errback.
import scrapy
from scrapy.crawler import CrawlerProcess
from scrapy.http.request import Request
class ProxySpider(scrapy.Spider):
name = "sslproxies"
check_url = "https://stackoverflow.com/questions/tagged/web-scraping"
proxy_link = "https://www.sslproxies.org/"
current_proxy = ""
proxies = {}
def start_requests(self):
yield Request(self.proxy_link,callback=self.parse_proxies)
def parse_proxies(self,response):
for row in response.css("table#proxylisttable tbody tr"):
if "yes" in row.extract():
td = row.css("td::text").extract()
self.proxies["http://{}".format(td[0]+":"+td[1])]={"valid":False}
for proxy in self.proxies.keys():
yield Request(self.check_url,callback=self.parse,errback=self.errback_httpbin,
meta={"proxy":proxy,
"download_slot":proxy},
dont_filter=True)
def parse(self, response):
if "proxy" in response.request.meta.keys():
#As script reaches this parse method we can mark current proxy as valid
self.proxies[response.request.meta["proxy"]]["valid"] = True
print(response.meta.get("proxy"))
if not self.current_proxy:
#Scraper reaches this code line on first valid response
self.current_proxy = response.request.meta["proxy"]
#yield Request(next_url, callback=self.parse_next,
# meta={"proxy":self.current_proxy,
# "download_slot":self.current_proxy})
def errback_httpbin(self, failure):
if "proxy" in failure.request.meta.keys():
proxy = failure.request.meta["proxy"]
if proxy == self.current_proxy:
#If current proxy after our usage becomes not valid
#Mark it as not valid
self.proxies[proxy]["valid"] = False
for ip_port in self.proxies.keys():
#And choose valid proxy from self.proxies
if self.proxies[ip_port]["valid"]:
failure.request.meta["proxy"] = ip_port
failure.request.meta["download_slot"] = ip_port
self.current_proxy = ip_port
return failure.request
print("Failure: "+str(failure))
if __name__ == "__main__":
c = CrawlerProcess({
'USER_AGENT': 'Mozilla/5.0',
'COOKIES_ENABLED': False,
'DOWNLOAD_TIMEOUT' : 10,
'DOWNLOAD_DELAY' : 3,
})
c.crawl(ProxySpider)
c.start()
you need write a downloader middleware, to install a process_exception hook, scrapy calls this hook when exception raised. in the hook, you could return a new Request object, with dont_filter=True flag, to let scrapy reschedule the request until it succeeds.
in the meanwhile, you could verify response extensively in process_response hook, check the status code, response content etc., and reschedule request as necessary.
in order to change proxy easily, you should use built-in HttpProxyMiddleware, instead of tinker with environ:
request.meta['proxy'] = proxy_address
take a look at this project as an example.
I'm trying to scrape a website for broken links, so far I have this code which is successfully logging in and crawling the site, but it's only recording HTTP status 200 codes:
class HttpStatusSpider(scrapy.Spider):
name = 'httpstatus'
handle_httpstatus_all = True
link_extractor = LinkExtractor()
def start_requests(self):
"""This method ensures we login before we begin spidering"""
# Little bit of magic to handle the CSRF protection on the login form
resp = requests.get('http://localhost:8000/login/')
tree = html.fromstring(resp.content)
csrf_token = tree.cssselect('input[name=csrfmiddlewaretoken]')[0].value
return [FormRequest('http://localhost:8000/login/', callback=self.parse,
formdata={'username': 'mischa_cs',
'password': 'letmein',
'csrfmiddlewaretoken': csrf_token},
cookies={'csrftoken': resp.cookies['csrftoken']})]
def parse(self, response):
item = HttpResponseItem()
item['url'] = response.url
item['status'] = response.status
item['referer'] = response.request.headers.get('Referer', '')
yield item
for link in self.link_extractor.extract_links(response):
r = Request(link.url, self.parse)
r.meta.update(link_text=link.text)
yield r
The docs and these answers lead me to believe that handle_httpstatus_all = True should cause scrapy to pass errored requests to my parse method, but so far I've not been able to capture any.
I've also experimented with handle_httpstatus_list and a custom errback handler in a different iteration of the code.
What do I need to change to capture the HTTP error codes scrapy is encountering?
handle_httpstatus_list can be defined on the spider level, but handle_httpstatus_all can only be defined on the Request level, including it on the meta argument.
I would still recommend using an errback for these cases, but if everything is controlled, it shouldn't create new problems.
So, I don't know if this is the proper scrapy way, but it does allow me to handle all HTTP status codes (including 5xx).
I disabled the HttpErrorMiddleware by adding this snippet to my scrapy project's settings.py:
SPIDER_MIDDLEWARES = {
'scrapy.spidermiddlewares.httperror.HttpErrorMiddleware': None
}
I'm using scrapy to crawl a web-site with authentication.
I want to be able to save the state of the crawler and I use
scrapy crawl myspider -s JOBDIR=mydir
After I resume with the same command I want to be able to login to the website before it reschedules all saved requests.
Basically, I want to be sure that my function login() and after_login() will be called before any other request is scheduled and executed. And I don't want to use cookies, because they don't allow me to pause the crawling for a long time.
I can call login() in start_requests(), but this works only when I run the crawler for the first time.
class MyCrawlSpider(CrawlSpider):
# ...
START_URLS = ['someurl.com', 'someurl2.com']
LOGIN_PAGE = u'https://login_page.php'
def login(self):
return Request(url=self.LOGIN_PAGE, callback=self.login_email,
dont_filter=True, priority=9999)
def login_form(self, response):
return FormRequest.from_response(response,
formdata={'Email': 'myemail',
'Passwd': 'mypasswd'},
callback=self.after_login,
dont_filter=True,
priority=9999)
def after_login(self, response):
if "authentication failed" in response.body:
self.logger.error("Login failed")
return
else:
print("Login Successful!!")
self.is_logged_in = True
for url in self.START_URLS:
yield Request(url, callback=self.parse_artists_json, dont_filter=True)
Bottomline: Is there any callback which will always be called when I resume crawling with -s JOBDIR=... option before rescheduling previous requests? I will use it to call login() method.
You can use the spider_opened signal (more here)
This function is intended to resource allocation for the spiders and others initializations, so it doesn't expect you to yield a Request object from there.
You can go around this by having an array of pending requests. This is needed because scrapy doesn't allow you to manually scheduled requests.
Then, after resume the spider you can queue the login as the first requests on the queue:
def spider_opened(self, spider):
self.spider.requests.insert(0, self.spider.login())
You also need to add a next_request method into your spider
def next_request(self):
if self.requests:
yield self.requests.pop(0)
And queue all you requests by adding them to the requests array, and calling next_request add the end of each method:
def after_login(self, response):
if "authentication failed" in response.body:
self.logger.error("Login failed")
return
else:
print("Login Successful!!")
self.is_logged_in = True
if not self.requests:
for url in self.START_URLS:
self.requests.append(Request(url, callback=self.parse_artists_json, dont_filter=True)
yield self.next_request()