I'm going through a set of pages and I'm not certain how many there are, but the current page is represented by a simple number present in the url (e.g. "http://www.website.com/page/1")
I would like to use a for loop in scrapy to increment the current guess at the page and stop when it reaches a 404. I know the response that is returned from the request contains this information, but I'm not sure how to automatically get a response from a request.
Any ideas on how to do this?
Currently my code is something along the lines of :
def start_requests(self):
baseUrl = "http://website.com/page/"
currentPage = 0
stillExists = True
while(stillExists):
currentUrl = baseUrl + str(currentPage)
test = Request(currentUrl)
if test.response.status != 404: #This is what I'm not sure of
yield test
currentPage += 1
else:
stillExists = False
You can do something like this:
from __future__ import print_function
import urllib2
baseURL = "http://www.website.com/page/"
for n in xrange(100):
fullURL = baseURL + str(n)
#print fullURL
try:
req = urllib2.Request(fullURL)
resp = urllib2.urlopen(req)
if resp.getcode() == 404:
#Do whatever you want if 404 is found
print ("404 Found!")
else:
#Do your normal stuff here if page is found.
print ("URL: {0} Response: {1}".format(fullURL, resp.getcode()))
except:
print ("Could not connect to URL: {0} ".format(fullURL))
This iterates through the range and attempts to connect to each URL via urllib2. I don't know scapy or how your example function opens the URL but this is an example with how to do it via urllib2.
Note that most sites that utilize this type of URL format are normally running a CMS that can automatically redirect non-existent pages to a custom 404 - Not Found page which will still show up as a HTTP status code of 200. In this case, the best way to look for a page that may show up but is actually just the custom 404 page, you should do some screen scraping and look for anything that may not appear during a "normal" page return such as text that says "Page not found" or something similar and unique to the custom 404 page.
You need to yield/return the request in order to check the status, creating a Request object does not actually send it.
class MySpider(BaseSpider):
name = 'website.com'
baseUrl = "http://website.com/page/"
def start_requests(self):
yield Request(self.baseUrl + '0')
def parse(self, response):
if response.status != 404:
page = response.meta.get('page', 0) + 1
return Request('%s%s' % (self.baseUrl, page), meta=dict(page=page))
Related
I am using twint for scraping twitter profiles.
When I run this script:
c = twint.Config()
c.Username = username
c.Store_object = True
c.Store_object_users_list = users
c.Hide_output = True
twint.run.Lookup(c)
try:
userna = users[0]
except:
continue
web = userna.url
I get the masked/shortened URL instead of a real one. How can I get the real url?
What would you advise?
Preface: This answer is based on my results and findings from evaluating UVuuMe's answer (in it's initial version).
To translate a shortened URL into the full URL it represents, you can use the package requests which comes indirectly with twint (it's required by googletransx which is required by twint which you already installed, so there is no need for pip install requests).
Send a HEAD request, then check the response's status_code for 303, only then read the location header; there are other cases where a location will be in the response, but not in case of HTTP 200 OK.
import requests
# Short URL for a Python Requests Tutorial
short_url = 'https://youtu.be/tb8gHvYlCFs'
res = requests.head(short_url)
if res.status_code == 303: # "See Other"
full_url = res.headers['location']
elif res.status_code == 200: # "OK"
# let's conclude that short_url is already what we are looking for
full_url = short_url
else:
# replace by your error handling:
assert(False)
The following works.
import requests
resp = requests.head(short_link)
resp.status_code
true_url = resp.headers["Location"]
i get data from an api in django.
The data comes from an order form from another website.
The data also includes an url, for example like example.com but i can't validate the input because i don't have access to the order form.
The url that i get can also have different kinds. More examples:
example.de
http://example.de
www.example.com
https://example.de
http://www.example.de
https://www.example.de
Now i would like to open the url to get the correct url.
For example if i open example.com in my browser, i got the correct url http://example.com/ and that is what i wish for all urls.
How can i do that in python fast?
If you get status_code 200 you know that you have a valid address.
In regards to HTTPS://. You will get an SSL error if you don't Follow the answers in this guide. Once you have that in place, the program will find the correct URL for you.
import requests
import traceback
validProtocols = ["https://www.", "http://www.", "https://", "http://"]
def removeAnyProtocol(url):
url = url.replace("www.","") # to remove any inputs containing just www since we aren't planning on using them regardless.
for protocol in validProtocols:
url = url.replace(protocol, "")
return url
def validateUrl(url):
for protocol in validProtocols:
if(protocol not in url):
pUrl = protocol + removeAnyProtocol(url)
try:
req = requests.head(pUrl, allow_redirects=True)
if req.status_code == 200:
return pUrl
else:
continue
except Exception:
print(traceback.format_exc())
continue
else:
try:
req = requests.head(url, allow_redirects=True)
if req.status_code == 200:
return url
except Exception:
print(traceback.format_exc())
continue
Usage:
correctUrl = validateUrl("google.com") # https://www.google.com
I've written a script in scrapy to make proxied requests using newly generated proxies by get_proxies() method. I used requests module to fetch the proxies in order to reuse them in the script. What I'm trying to do is parse all the movie links from it's landing page and then fetch the name of each movie from it's target page. My following script can use rotation of proxies.
I know there is an easier way to change proxies, like it is described here HttpProxyMiddleware but I would still like to stick to the way I'm trying here.
website link
This is my current attempt (It keeps using new proxies to fetch a valid response but every time it gets 503 Service Unavailable):
import scrapy
import random
import requests
from itertools import cycle
from bs4 import BeautifulSoup
from scrapy.crawler import CrawlerProcess
def get_proxies():
response = requests.get("https://www.us-proxy.org/")
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
class ProxySpider(scrapy.Spider):
name = "proxiedscript"
handle_httpstatus_list = [503]
proxy_vault = get_proxies()
check_url = "https://yts.am/browse-movies"
def start_requests(self):
random.shuffle(self.proxy_vault)
proxy_url = next(cycle(self.proxy_vault))
request = scrapy.Request(self.check_url,callback=self.parse,dont_filter=True)
request.meta['https_proxy'] = f'http://{proxy_url}'
yield request
def parse(self,response):
print(response.meta)
if "DDoS protection by Cloudflare" in response.css(".attribution > a::text").get():
random.shuffle(self.proxy_vault)
proxy_url = next(cycle(self.proxy_vault))
request = scrapy.Request(self.check_url,callback=self.parse,dont_filter=True)
request.meta['https_proxy'] = f'http://{proxy_url}'
yield request
else:
for item in response.css(".browse-movie-wrap a.browse-movie-title::attr(href)").getall():
nlink = response.urljoin(item)
yield scrapy.Request(nlink,callback=self.parse_details)
def parse_details(self,response):
name = response.css("#movie-info h1::text").get()
yield {"Name":name}
if __name__ == "__main__":
c = CrawlerProcess({'USER_AGENT':'Mozilla/5.0'})
c.crawl(ProxySpider)
c.start()
To make sure whether the request is being proxied, I printed response.meta and could get results like this {'https_proxy': 'http://142.93.127.126:3128', 'download_timeout': 180.0, 'download_slot': 'yts.am', 'download_latency': 0.237013578414917, 'retry_times': 2, 'depth': 0}.
As I've overused the link to check how the proxied request within scrapy works, I'm getting 503 Service Unavailable error at this moment and I can see this keyword within the response DDoS protection by Cloudflare. However, I get valid response when I try with requests module applying the same logic I implemented here.
My earlier question: why I can't get the valid response as (I suppose) I'm using proxies in the right way? [solved]
Bounty Question: how can I define try/except clause within my script so that it will try with different proxies once it throws connection error with a certain proxy?
According to scrapy.downloadermiddlewares.httpproxy.HttpProxyMiddleware docs
(and source)
proxy meta key is expected to use (not https_proxy)
#request.meta['https_proxy'] = f'http://{proxy_url}'
request.meta['proxy'] = f'http://{proxy_url}'
As scrapy didn't received valid meta key - your scrapy application didn't use proxies
The start_requests() function is just the entry point. On subsequent requests, you would need to resupply this metadata to the Request object.
Also, errors can occur on two levels: proxy and target server
We need to handle bad response codes from both the proxy and the target server. Proxy errors are returned by the middelware to the errback function. The target server response can be handled during parsing from the response.status
import scrapy
import random
import requests
from itertools import cycle
from bs4 import BeautifulSoup
from scrapy.crawler import CrawlerProcess
def get_proxies():
response = requests.get("https://www.us-proxy.org/")
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]
# proxy = ['https://52.0.0.1:8090', 'https://52.0.0.2:8090']
return proxy
def get_random_proxy(proxy_vault):
random.shuffle(proxy_vault)
proxy_url = next(cycle(proxy_vault))
return proxy_url
class ProxySpider(scrapy.Spider):
name = "proxiedscript"
handle_httpstatus_list = [503, 502, 401, 403]
check_url = "https://yts.am/browse-movies"
proxy_vault = get_proxies()
def handle_middleware_errors(self, *args, **kwargs):
# implement middleware error handling here
print('Middleware Error')
# retry request with different proxy
yield self.make_request(url=args[0].request._url, callback=args[0].request._meta['callback'])
def start_requests(self):
yield self.make_request(url=self.check_url, callback=self.parse)
def make_request(self, url, callback, dont_filter=True):
return scrapy.Request(url,
meta={'proxy': f'https://{get_random_proxy(self.proxy_vault)}', 'callback': callback},
callback=callback,
dont_filter=dont_filter,
errback=self.handle_middleware_errors)
def parse(self, response):
print(response.meta)
try:
if response.status != 200:
# implement server status code handling here - this loops forever
print(f'Status code: {response.status}')
raise
else:
for item in response.css(".browse-movie-wrap a.browse-movie-title::attr(href)").getall():
nlink = response.urljoin(item)
yield self.make_request(url=nlink, callback=self.parse_details)
except:
# if anything goes wrong fetching the lister page, try again
yield self.make_request(url=self.check_url, callback=self.parse)
def parse_details(self, response):
print(response.meta)
try:
if response.status != 200:
# implement server status code handeling here - this loops forever
print(f'Status code: {response.status}')
raise
name = response.css("#movie-info h1::text").get()
yield {"Name": name}
except:
# if anything goes wrong fetching the detail page, try again
yield self.make_request(url=response.request._url, callback=self.parse_details)
if __name__ == "__main__":
c = CrawlerProcess({'USER_AGENT': 'Mozilla/5.0'})
c.crawl(ProxySpider)
c.start()
I've been looking through the Python Requests documentation but I cannot see any functionality for what I am trying to achieve.
In my script I am setting allow_redirects=True.
I would like to know if the page has been redirected to something else, what is the new URL.
For example, if the start URL was: www.google.com/redirect
And the final URL is www.google.co.uk/redirected
How do I get that URL?
You are looking for the request history.
The response.history attribute is a list of responses that led to the final URL, which can be found in response.url.
response = requests.get(someurl)
if response.history:
print("Request was redirected")
for resp in response.history:
print(resp.status_code, resp.url)
print("Final destination:")
print(response.status_code, response.url)
else:
print("Request was not redirected")
Demo:
>>> import requests
>>> response = requests.get('http://httpbin.org/redirect/3')
>>> response.history
(<Response [302]>, <Response [302]>, <Response [302]>)
>>> for resp in response.history:
... print(resp.status_code, resp.url)
...
302 http://httpbin.org/redirect/3
302 http://httpbin.org/redirect/2
302 http://httpbin.org/redirect/1
>>> print(response.status_code, response.url)
200 http://httpbin.org/get
This is answering a slightly different question, but since I got stuck on this myself, I hope it might be useful for someone else.
If you want to use allow_redirects=False and get directly to the first redirect object, rather than following a chain of them, and you just want to get the redirect location directly out of the 302 response object, then r.url won't work. Instead, it's the "Location" header:
r = requests.get('http://github.com/', allow_redirects=False)
r.status_code # 302
r.url # http://github.com, not https.
r.headers['Location'] # https://github.com/ -- the redirect destination
I think requests.head instead of requests.get will be more safe to call when handling url redirect. Check a GitHub issue here:
r = requests.head(url, allow_redirects=True)
print(r.url)
the documentation has this blurb https://requests.readthedocs.io/en/master/user/quickstart/#redirection-and-history
import requests
r = requests.get('http://www.github.com')
r.url
#returns https://www.github.com instead of the http page you asked for
For python3.5, you can use the following code:
import urllib.request
res = urllib.request.urlopen(starturl)
finalurl = res.geturl()
print(finalurl)
I wrote the following function to get the full URL from a short URL (bit.ly, t.co, ...)
import requests
def expand_short_url(url):
r = requests.head(url, allow_redirects=False)
r.raise_for_status()
if 300 < r.status_code < 400:
url = r.headers.get('Location', url)
return url
Usage (short URL is this question's url):
short_url = 'https://tinyurl.com/' + '4d4ytpbx'
full_url = expand_short_url(short_url)
print(full_url)
Output:
https://stackoverflow.com/questions/20475552/python-requests-library-redirect-new-url
I wasn't able to use requests library and had to go different way. Here is the code that I post as solution to this post. (To get redirected URL with requests)
This way you actually open the browser, wait for your browser to log the url in the history log and then read last url in your history. I wrote this code for google chrom, but you should be able to follow along if you are using different browser.
import webbrowser
import sqlite3
import pandas as pd
import shutil
webbrowser.open("https://twitter.com/i/user/2274951674")
#source file is where the history of your webbroser is saved, I was using chrome, but it should be the same process if you are using different browser
source_file = 'C:\\Users\\{your_user_id}\\AppData\\Local\\Google\\Chrome\\User Data\\Default\\History'
# could not directly connect to history file as it was locked and had to make a copy of it in different location
destination_file = 'C:\\Users\\{user}\\Downloads\\History'
time.sleep(30) # there is some delay to update the history file, so 30 sec wait give it enough time to make sure your last url get logged
shutil.copy(source_file,destination_file) # copying the file.
con = sqlite3.connect('C:\\Users\\{user}\\Downloads\\History')#connecting to browser history
cursor = con.execute("SELECT * FROM urls")
names = [description[0] for description in cursor.description]
urls = cursor.fetchall()
con.close()
df_history = pd.DataFrame(urls,columns=names)
last_url = df_history.loc[len(df_history)-1,'url']
print(last_url)
>>https://twitter.com/ozanbayram01
All the answers are applicable where the final url exists/working fine.
In case, final URL doesn't seems to work then below is way to capture all redirects.
There was scenario where final URL isn't working anymore and other ways like url history give error.
Code Snippet
long_url = ''
url = 'http://example.com/bla-bla'
try:
while True:
long_url = requests.head(url).headers['location']
print(long_url)
url = long_url
except:
print(long_url)
I have followed several other questions in SO to find the final redirect url, however for the following url I can't make the redirect work. It doesn't redirect and stays at tinyurl.
import urllib2
def getFinalUrl(start_url):
var = urllib2.urlopen(start_url)
final_url = var.geturl()
return final_url
url = "http://redirect.tinyurl.com/api/click?key=a7e37b5f6ff1de9cb410158b1013e54a&out=http%3A%2F%2Fwww.amazon.com%2Fgp%2Fprofile%2FA3B4EO22KUPKYW&loc=&cuid=0072ce987ebb47328d22e465a051ce7&opt=false&format=txt"
redirect = getFinalUrl(url)
print "redirect: " + redirect
the result (which is not the final url if you try in a browser):
redirect: http://redirect.tinyurl.com/api/click?key=a7e37b5f6ff1de9cb410158b1013e54a&out=http%3A%2F%2Fwww.amazon.com%2Fgp%2Fprofile%2FA3B4EO22KUPKYW&loc=&cuid=0072ce987ebb47328d22e465a051ce7&opt=false&format=txt
import urlparse
url = 'http://redirect.tinyurl.com/api/click?key=a7e37b5f6ff1de9cb410158b1013e54a&out=http%3A%2F%2Fwww.amazon.com%2Fgp%2Fprofile%2FA3B4EO22KUPKYW&loc=&cuid=0072ce987ebb47328d22e465a051ce7&opt=false&format=txt'
try:
out = urlparse.parse_qs(urlparse.urlparse(url).query)['out'][0]
print(out) #http://www.amazon.com/gp/profile/A3B4EO22KUPKYW
except Exception as e: # dont catch all
print('not found')
This kind of url does not need to be curled to find out what the destination/redirect url is, well, because you ALREADY have them in your url.
If the destination/redirect url is not shown like this guy
tinyurl.com/xxxx
then that's a different story, you'd have to curl it to find out what it resolves/304 to like below:
import requests
url = 'http://urlshortener.com/applebanana'
t = requests.get(url)
print(t.url)