I'm working on a Scrapy app, where I'm trying to login to a site with a form that uses a captcha (It's not spam). I am using ImagesPipeline to download the captcha, and I am printing it to the screen for the user to solve. So far so good.
My question is how can I restart the spider, to submit the captcha/form information? Right now my spider requests the captcha page, then returns an Item containing the image_url of the captcha. This is then processed/downloaded by the ImagesPipeline, and displayed to the user. I'm unclear how I can resume the spider's progress, and pass the solved captcha and same session to the spider, as I believe the spider has to return the item (e.g. quit) before the ImagesPipeline goes to work.
I've looked through the docs and examples but I haven't found any ones that make it clear how to make this happen.
This is how you might get it to work inside the spider.
self.crawler.engine.pause()
process_my_captcha()
self.crawler.engine.unpause()
Once you get the request, pause the engine, display the image, read the info from the user& resume the crawl by submitting a POST request for login.
I'd be interested to know if the approach works for your case.
I would not create an Item and use the ImagePipeline.
import urllib
import os
import subprocess
...
def start_requests(self):
request = Request("http://webpagewithcaptchalogin.com/", callback=self.fill_login_form)
return [request]
def fill_login_form(self,response):
x = HtmlXPathSelector(response)
img_src = x.select("//img/#src").extract()
#delete the captcha file and use urllib to write it to disk
os.remove("c:\captcha.jpg")
urllib.urlretrieve(img_src[0], "c:\captcha.jpg")
# I use an program here to show the jpg (actually send it somewhere)
captcha = subprocess.check_output(r".\external_utility_solving_captcha.exe")
# OR just get the input from the user from stdin
captcha = raw_input("put captcha in manually>")
# this function performs the request and calls the process_home_page with
# the response (this way you can chain pages from start_requests() to parse()
return [FormRequest.from_response(response,formnumber=0,formdata={'user':'xxx','pass':'xxx','captcha':captcha},callback=self.process_home_page)]
def process_home_page(self, response):
# check if you logged in etc. etc.
...
What I do here is that I import urllib.urlretrieve(url) (to store the image), os.remove(file) (to delete the previous image), and subprocess.checoutput (to call an external command line utility to solve the captcha). The whole Scrapy infrastructure is not used in this "hack", because solving a captcha like this is always a hack.
That whole calling external subprocess thing could have been one nicer, but this works.
On some sites it's not possible to save the captcha image and you have to call the page in a browser and call a screen_capture utility and crop on an exact location to "cut out" the captcha. Now that is screenscraping.
Related
I'm trying to save screenshots of scraped webpages with Scrapy Splash. I've copied and pasted the code found here into my pipeline folder: https://docs.scrapy.org/en/latest/topics/item-pipeline.html
Here's the code from the url:
import scrapy
import hashlib
from urllib.parse import quote
class ScreenshotPipeline(object):
"""Pipeline that uses Splash to render screenshot of
every Scrapy item."""
SPLASH_URL = "http://localhost:8050/render.png?url={}"
async def process_item(self, item, spider):
encoded_item_url = quote(item["url"])
screenshot_url = self.SPLASH_URL.format(encoded_item_url)
request = scrapy.Request(screenshot_url)
response = await spider.crawler.engine.download(request, spider)
if response.status != 200:
# Error happened, return item.
return item
# Save screenshot to file, filename will be hash of url.
url = item["url"]
url_hash = hashlib.md5(url.encode("utf8")).hexdigest()
filename = "{}.png".format(url_hash)
with open(filename, "wb") as f:
f.write(response.body)
# Store filename in item.
item["screenshot_filename"] = filename
return item
I've also followed the instructions for setting up splash found here: https://github.com/scrapy-plugins/scrapy-splash
When I call the command scrapy crawl spidereverything works correctly except the pipeline.
This is the "Error" I'm seeing.
<coroutine object ScreenshotPipeline.process_item at 0x7f29a9c7c8c0>
The spider is yielding the item correctly, but it will not process the item.
Does anyone have any advice? Thank you.
Edit:
I think what is going on is that Scrapy is calling the process_item() method as you normally would. However according to these docs: https://docs.python.org/3/library/asyncio-task.html a coroutine object must be called differently.
asyncio.run(process_item()) rather than process_item().
I think I may have to modify the source code?
You should use scrapy-splash in a script inside spider not in the pipelines.
I followed this docs and it works for me.
I am working on a project that is divided into two parts:
Retrieve a specific page
Once the ID of this page is extracted,
Send requests to an API to obtain additional information on this page
For the second point, and to follow Scrapy's asynchronous philosophy, where should such a code be placed? (I hesitate between in the spider or in a pipeline).
Do we have to use different libraries like asyncio & aiohttp to be able to achieve this goal asynchronously? (I love aiohttp so this is not a problem to use it)
Thanks you
Since you're doing this to fetch additional information about an item, I'd just yield a request from the parsing method, passing the already scraped information in the meta attribute.
You can see an example of this at https://doc.scrapy.org/en/latest/topics/request-response.html#topics-request-response-ref-request-callback-arguments
This can also be done in a pipeline (either using scrapy's engine API, or a different library, e.g. treq).
I do however think that doing it "the normal way" from the spider makes more sense in this instance.
I recently had the same problem (again) and found an elegant solution using Twisted decorators t.i.d.inlineCallbacks.
# -*- coding: utf-8 -*-
import scrapy
import re
from twisted.internet.defer import inlineCallbacks
from sherlock import utils, items, regex
class PagesSpider(scrapy.spiders.SitemapSpider):
name = 'pages'
allowed_domains = ['thing.com']
sitemap_follow = [r'sitemap_page']
def __init__(self, site=None, *args, **kwargs):
super(PagesSpider, self).__init__(*args, **kwargs)
#inlineCallbacks
def parse(self, response):
# things
request = scrapy.Request("https://google.com")
response = yield self.crawler.engine.download(request, self)
# Twisted execute the request and resume the generator here with the response
print(response.text)
This is the first time I ask question here. If something I got wrong, please forgive me.
And I am a newer in python for one month, I try to use the scrapy to learn something more about spider.
question is here:
def get_chapterurl(self, response):
item = DingdianItem()
item['name'] = str(response.meta['name']).replace('\xa0', '')
yield item
yield Request(url=response.url, callback=self.get_chapter, meta={'name':name_id})
def get_chapter(self, response):
urls = re.findall(r'<td class="L">(.*?)</td>', response.text)
As you can see, I yield item and Requests at the same time, but the get_chapter function did not run the first line(I take a break point there), so where was I wrong?
Sorry for disturbing you.
I have google for a time, but get noting...
Your request gets filtered out.
Scrapy has in-built request filter that prevents you from downloading the same page twice (intended feature).
Lets say you are on http://example.com; this request you yield:
yield Request(url=response.url, callback=self.get_chapter, meta={'name':name_id})
tries to download http://example.com again. And if you look at the crawling log it should say something along the lines of "ignoring duplicate url http://example.com".
You can always ignore this feature by setting dont_filter=True parameter in your Request object, as so:
yield Request(url=response.url, callback=self.get_chapter, meta={'name':name_id},
dont_filter=True)
However! I'm having trouble understanding the intention of your code but it seems that you don't really want to download the same url twice.
You don't have to schedule a new request either, you can just call your callback with the request you already have:
response = response.replace(meta={'name': name_id}) # update meta
# why crawl it again, if we can just call the callback directly!
# for python2
for result in self.get_chapter(response):
yield result
# or if you are running python3:
yield from self.get_chapter(response):
The following django middleware function is used to identify active links in a django response object. If a link is active, its marked with a css class and the href attribute gets replaced by javascript:void(null);. Using this function, the last two lines before the return statement are so slow that i cant use it, further, no css, js and images are rendered. However, if i put these two calls into the for loop, everything is fine and fast. But, i dont want these two calls executed for each active link on page, instead i want them executed only once, and that doesnt work, i really cant see why and what the for loop has to do with it. Its no BeautifulSoup issue, because its the same with re.sub('\s+','',response.content) or the replace function. As far as i have investigated this, i can tell you that the very last line before the return statement is the slow one, as long as its not executed inside the for loop. I'm really excited about a possible explanation.
import re
from django_projects.projects.my_project.settings import SITE_NAME
from BeautifulSoup import BeautifulSoup
class PostRender():
def process_response(self, request, response):
link_pattern=re.compile('<a.*href="(http://%s)*%s".*>' % (SITE_NAME,request.path),re.IGNORECASE)
klass_pattern=re.compile('class="[^"]*"',re.IGNORECASE)
href_pattern=re.compile('href="(http://%s)*%s(\?.*)*"' % (SITE_NAME,request.path),re.IGNORECASE)
#find all active links
search=re.finditer(link_pattern ,response.content)
for res in search:
result=res.group()
klassname='class="active"'
if 'class' in result:
klass=re.search(klass_pattern,result).group().split('=')[1]
if len(klass) != 0:
klassname='class="%s %s"' % (klass[1:-1],'active')
link=re.sub(href_pattern,'href="javascript:void(null);"',re.sub(klass_pattern,klassname,result))
response.content=re.sub(result,link,response.content)
soup=BeautifulSoup(response.content)
response.content=soup.prettify()
return response
I have some JavaScript from a 3rd party vendor that is initiating an image request. I would like to figure out the URI of this image request.
I can load the page in my browser, and then monitor "Live HTTP Headers" or "Tamper Data" in order to figure out the image request URI, but I would prefer to create a command line process to do this.
My intuition is that it might be possible using python + qtwebkit, but perhaps there is a better way.
To clarify: I might have this (overly simplified code).
<script>
suffix = magicNumberFunctionIDontHaveAccessTo();
url = "http://foobar.com/function?parameter=" + suffix
img = document.createElement('img'); img.src=url; document.all.body.appendChild(img);
</script>
Then once the page is loaded, I can go figure out the url by sniffing the packets. But I can't just figure it out from the source, because I can't predict the outcome of magicNumberFunction...().
Any help would be muchly appreciated!
Thank you.
The simplest thing to do might be to use something like HtmlUnit and skip a real browser entirely. By using Rhino, it can evaluate JavaScript and likely be used to extract that URL out.
That said, if you can't get that working, try out Selenium RC and use the captureNetworkTraffic command (which requires the Selenium instant be started with an option of captureNetworkTraffic=true). This will launch Firefox with a proxy configured and then let you pull the request info back out as JSON/XML/plain text. Then you can parse that content and get what you want.
Try out the instant test tool that my company offers. If the data you're looking for is in our results (after you click View Details), you'll be able to get it from Selenium. I know, since I wrote the captureNetworkTraffic API for Selenium for my company, BrowserMob.
I would pick any one of the many http proxy servers written in Python -- probably one of the simplest ones at the very top of the list -- and tweak it to record all URLs requested (as well as proxy-serve them) e.g. appending them to a text file -- without loss of generality, call that text file 'XXX.txt'.
Now all you need is a script that: starts the proxy server in question; starts Firefox (or whatever) on your main desired URL with the proxy in question set as your proxy (see e.g. this SO question for how), though I'm sure other browsers would work just as well; waits a bit (e.g. until the proxy's XXX.txt file has not been altered for more than N seconds); reads XXX.txt to extract only the URLs you care about and record them wherever you wish; turns down the proxy and Firefox processes.
I think this will be much faster to put in place and make work correctly, for your specific requirements, than any more general solution based on qtwebkit, selenium, or other "automation kits".
Use Firebug Firefox plugin. It will show you all requests in real time and you can even debug the JS in your Browser or run it step-by-step.
Ultimately, I did it in python, using Selenium-RC. This solution requires the python files for selenium-rc, and you need to start the java server ("java -jar selenium-server.jar")
from selenium import selenium
import unittest
import lxml.html
class TestMyDomain(unittest.TestCase):
def setUp(self):
self.selenium = selenium("localhost", \
4444, "*firefox", "http://www.MyDomain.com")
self.selenium.start()
def test_mydomain(self):
htmldoc = open('site-list.html').read()
url_list = [link for (element, attribute,link,pos) in lxml.html.iterlinks(htmldoc)]
for url in url_list:
try:
sel = self.selenium
sel.open(url)
sel.select_window("null")
js_code = '''
myDomainWindow = this.browserbot.getUserWindow();
for(obj in myDomainWindow) {
/* This code grabs the OMNITURE tracking pixel img */
if ((obj.substring(0,4) == 's_i_') && (myDomainWindow[obj].src)) {
var ret = myDomainWindow[obj].src;
}
}
ret;
'''
omniture_url = sel.get_eval(js_code) #parse&process this however you want
except Exception, e:
print 'We ran into an error: %s' % (e,)
self.assertEqual("expectedValue", observedValue)
def tearDown(self):
self.selenium.stop()
if __name__ == "__main__":
unittest.main()
Why can't you just read suffix, or url for that matter? Is the image loaded in an iframe or in your page?
If it is loaded in your page, then this may be a dirty hack (substitute document.body for whatever element is considered):
var ac = document.body.appendChild;
var sources = [];
document.body.appendChild = function(child) {
if (/^img$/i.test(child.tagName)) {
sources.push(child.getAttribute('src'));
}
ac(child);
}