I am using Python request library to scrape robots.txt data from a list of URLs:
for url in urls:
url = urllib.parse.urljoin(url, "robots.txt")
try:
r = requests.get(url, headers=headers, allow_redirects=False)
r.raise_for_status()
extract_robots(r)
except (exceptions.RequestException, exceptions.HTTPError, exceptions.Timeout) as err:
handle_exeption(err)
In my list of urls, I have this webpage: https://reward.ff.garena.com. When I am requesting https://reward.ff.garena.com/robots.txt, I am directly redirected to https://reward.ff.garena.com/en. However, I specified in my request parameters that I don't want redirects allow_redirects=False.
How can I skip this kind of redirect and make sure I only have domain/robots.txt data calling my extract_robots(data) method?
Do you know for sure that there is a robots.txt at that location?
I note that if I request https://reward.ff.garena.com/NOSUCHFILE.txt that I get the same result as for robots.txt
The allow_redirects=False only stops requests from automatically following 302/location= responses - i.e. it doesn’t stop the server you’re trying to access from returning a redirect as the response to the request you’re making.
If you get this type of response I guess it indicates the file you requested isn’t available, or some other error preventing you accessing it, perhaps in the general case of file access this might indicate need for authentication but for robots.txt that shouldn’t be the problem - simplest to assume the robots.txt isn’t there.
I'm trying to check multiple URLs on Google Safebrowsing API, but it returns an empty response every time. Have been googling for quite few hours with no results, and I don't need some overkill library for a simple POST request.
Edit: Using Python 3.5.2
import requests
import json
api_key = '123456'
url = "https://safebrowsing.googleapis.com/v4/threatMatches:find"
payload = {'client': {'clientId': "mycompany", 'clientVersion': "0.1"},
'threatInfo': {'threatTypes': ["SOCIAL_ENGINEERING", "MALWARE"],
'platformTypes': ["ANY_PLATFORM"],
'threatEntryTypes': ["URL"],
'threatEntries': [{'url': "http://www.thetesturl.com"}]}}
params = {'key': api_key}
r = requests.post(url, params=params, json=payload)
# Print response
print(r)
print(r.json())
This is my code, that returns HTTP 200 OK, but the response is empty.
What am I doing wrong?
I have the feeling that the api is not working properly. It returns 200 empty result even for urls that are marked as dangerous. For example, I checked this url using google's form and got the result `Some pages on this site are unsafe``. But using the api, it returns 200 empty... I believe it returns results only for specific pages. If only some pages are infected/dangerous, then you won't get any data for the main domain... Not very useful if you're asking me, but hey...it's free.
It would be nice if someone from Google could confirm this and add it to the documentation.
A real malware test url would have been much appreciated also, so we can test with real data.
According to the Safe Browsing API documentation, if you receive an empty object is because there was no match found:
Note: If there are no matches (that is, if none of the URLs specified
in the request are found on any of the lists specified in a request),
the HTTP POST response simply returns an empty object in the response
body.
I would like to scrape just the title of a webpage using Python. I need to do this for thousands of sites so it has to be fast. I've seen previous questions like retrieving just the title of a webpage in python, but all of the ones I've found download the entire page before retrieving the title, which seems highly inefficient as most often the title is contained within the first few lines of HTML.
Is it possible to download only the parts of the webpage until the title has been found?
I've tried the following, but page.readline() downloads the entire page.
import urllib2
print("Looking up {}".format(link))
hdr = {'User-Agent': 'Mozilla/5.0',
'Accept': 'text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8',
'Accept-Charset': 'ISO-8859-1,utf-8;q=0.7,*;q=0.3',
'Accept-Encoding': 'none',
'Accept-Language': 'en-US,en;q=0.8',
'Connection': 'keep-alive'}
req = urllib2.Request(link, headers=hdr)
page = urllib2.urlopen(req, timeout=10)
content = ''
while '</title>' not in content:
content = content + page.readline()
-- Edit --
Note that my current solution makes use of BeautifulSoup constrained to only process the title so the only place I can optimize is likely to not read in the entire page.
title_selector = SoupStrainer('title')
soup = BeautifulSoup(page, "lxml", parse_only=title_selector)
title = soup.title.string.strip()
-- Edit 2 --
I've found that BeautifulSoup itself splits the content into multiple strings in the self.current_data
variable (see this function in bs4), but I'm unsure how to modify the code to basically stop reading all remaining content after the title has been found. One issue could be that redirects should still work.
-- Edit 3 --
So here's an example. I have a link www.xyz.com/abc and I have to follow this through any redirects (almost all of my links use a bit.ly kind of link shortening). I'm interested in both the title and domain that occurs after any redirections.
-- Edit 4 --
Thanks a lot for all of your assistance! The answer by Kul-Tigin works very well and has been accepted. I'll keep the bounty until it runs out though to see if a better answer comes up (as shown by e.g. a time measurement comparison).
-- Edit 5 --
For anyone interested: I've timed the accepted answer to be roughly twice as fast as my existing solution using BeautifulSoup4.
You can defer downloading the entire response body by enabling stream mode of requests.
Requests 2.14.2 documentation - Advanced Usage
By default, when you make a request, the body of the response is
downloaded immediately. You can override this behaviour and defer
downloading the response body until you access the Response.content
attribute with the stream parameter:
...
If you set stream to True when making a request, Requests cannot release the connection back to the pool unless you consume all the data or call Response.close.
This can lead to inefficiency with connections. If you find yourself partially reading request bodies (or not reading them at all) while using stream=True, you should consider using contextlib.closing (documented here)
So, with this method, you can read the response chunk by chunk until you encounter the title tag. Since the redirects will be handled by the library you'll be ready to go.
Here's an error-prone code tested with Python 2.7.10 and 3.6.0:
try:
from HTMLParser import HTMLParser
except ImportError:
from html.parser import HTMLParser
import requests, re
from contextlib import closing
CHUNKSIZE = 1024
retitle = re.compile("<title[^>]*>(.*?)</title>", re.IGNORECASE | re.DOTALL)
buffer = ""
htmlp = HTMLParser()
with closing(requests.get("http://example.com/abc", stream=True)) as res:
for chunk in res.iter_content(chunk_size=CHUNKSIZE, decode_unicode=True):
buffer = "".join([buffer, chunk])
match = retitle.search(buffer)
if match:
print(htmlp.unescape(match.group(1)))
break
Question: ... the only place I can optimize is likely to not read in the entire page.
This does not read the entire page.
Note: Unicode .decode() will raise Exception if you cut a Unicode sequence in the middle. Using .decode(errors='ignore') remove those sequences.
For instance:
import re
try:
# PY3
from urllib import request
except:
import urllib2 as request
for url in ['http://www.python.org/', 'http://www.google.com', 'http://www.bit.ly']:
f = request.urlopen(url)
re_obj = re.compile(r'.*(<head.*<title.*?>(.*)</title>.*</head>)',re.DOTALL)
Found = False
data = ''
while True:
b_data = f.read(4096)
if not b_data: break
data += b_data.decode(errors='ignore')
match = re_obj.match(data)
if match:
Found = True
title = match.groups()[1]
print('title={}'.format(title))
break
f.close()
Output:
title=Welcome to Python.org
title=Google
title=Bitly | URL Shortener and Link Management Platform
Tested with Python: 3.4.2 and 2.7.9
You're scraping webpages using standard REST requests and I'm not aware of any request that only returns the title, so I don't think it's possible.
I know this doesn't necessarily help get the title only, but I usually use BeautifulSoup for any web scraping. It's much easier. Here's an example.
Code:
import requests
from bs4 import BeautifulSoup
urls = ["http://www.google.com", "http://www.msn.com"]
for url in urls:
r = requests.get(url)
soup = BeautifulSoup(r.text, "html.parser")
print "Title with tags: %s" % soup.title
print "Title: %s" % soup.title.text
print
Output:
Title with tags: <title>Google</title>
Title: Google
Title with tags: <title>MSN.com - Hotmail, Outlook, Skype, Bing, Latest News, Photos & Videos</title>
Title: MSN.com - Hotmail, Outlook, Skype, Bing, Latest News, Photos & Videos
the kind of thing you want i don't think can be done, since the way the web is set up, you get the response for a request before anything is parsed. there isn't usually a streaming "if encounter <title> then stop giving me data" flag. if there is id love to see it, but there is something that may be able to help you. keep in mind, not all sites respect this. so some sites will force you to download the entire page source before you can act on it. but a lot of them will allow you to specify a range header. so in a requests example:
import requests
targeturl = "http://www.urbandictionary.com/define.php?term=Blarg&page=2"
rangeheader = {"Range": "bytes=0-150"}
response = requests.get(targeturl, headers=rangeheader)
response.text
and you get
'<!DOCTYPE html>\n<html lang="en-US" prefix="og: http://ogp.me/ns#'
now of course here's the problems with this
what if you specify a range that is too short to get the title of the page?
whats a good range to aim for? (combination of speed and assurance of accuracy)
what happens if the page doesn't respect Range? (most of the time you just get the whole response you would have without it.)
i don't know if this might help you? i hope so. but i've done similar things to only get file headers for download checking.
EDIT4:
so i thought of another kind of hacky thing that might help. nearly every page has a 404 page not found page. we might be able to use this to our advantage. instead of requesting the regular page. request something like this.
http://www.urbandictionary.com/nothing.php
the general page will have tons of information, links, data. but the 404 page is nothing more than a message, and (in this case) a video. and usually there is no video. just some text.
but you also notice that the title still appears here. so perhaps we can just request something we know does not exist on any page like.
X5ijsuUJSoisjHJFk948.php
and get a 404 for each page. that way you only download a very small and minimalistic page. nothing more. which will significantly reduce the amount of information you download. thus increasing speed and efficiency.
heres the problem with this method: you need to check somehow if the page does not supply its own version of the 404. most pages have it because it looks good with the site. and its standard practice to include one. but not all of them do. make sure to handle this case.
but i think that could be something worth trying out. over the course of thousands of sites, it would save many ms of download time for each html.
EDIT5:
so as we talked about, since you are interested in urls that redirect. we might make use of an http head reqeust. which wont get the site content. just the headers. so in this case:
response = requests.head('http://myshortenedurl.com/5b2su2')
replace my shortenedurl with tunyurl to follow along.
>>>response
<Response [301]>
nice so we know this redirects to something.
>>>response.headers['Location']
'http://stackoverflow.com'
now we know where the url redirects to without actually following it or downloading any page source. now we can apply any of the other techniques previously discussed.
Heres an example, using requests and lxml modules and using the 404 page idea. (be aware, i have to replace bit.ly with bit'ly so stack overflow doesnt get mad.)
#!/usr/bin/python3
import requests
from lxml.html import fromstring
links = ['http://bit'ly/MW2qgH',
'http://bit'ly/1x0885j',
'http://bit'ly/IFHzvO',
'http://bit'ly/1PwR9xM']
for link in links:
response = '<Response [301]>'
redirect = ''
while response == '<Response [301]>':
response = requests.head(link)
try:
redirect = response.headers['Location']
except Exception as e:
pass
fakepage = redirect + 'X5ijsuUJSoisjHJFk948.php'
scrapetarget = requests.get(fakepage)
tree = fromstring(scrapetarget.text)
print(tree.findtext('.//title'))
so here we get the 404 pages, and it will follow any number of redirects. now heres the output from this:
Urban Dictionary error
Page Not Found - Stack Overflow
Error 404 (Not Found)!!1
Kijiji: Page Not Found
so as you can see we did indeed get out titles. but we see some problems with the method. namely some titles add things, and some just dont have a good title at all. and thats the issue with that method. we could however try the range method too. benefits of that would be the title would be correct, but sometimes we might miss it, and sometimes we have to download the whole pagesource to get it. increasing required time.
Also credit to alecxe for this part of my quick and dirty script
tree = fromstring(scrapetarget.text)
print(tree.findtext('.//title'))
for an example with the range method. in the loop for link in links: change the code after the try catch statement to this:
rangeheader = {"Range": "bytes=0-500"}
scrapetargetsection = requests.get(redirect, headers=rangeheader)
tree = fromstring(scrapetargetsection.text)
print(tree.findtext('.//title'))
output is:
None
Stack Overflow
Google
Kijiji: Free Classifieds in...
here we see urban dictionary has no title or ive missed it in the bytes returned. in any of these methods there are tradeoffs. the only way to get close to total accuracy would be to download the entire source for each page i think.
using urllib you can set the Range header to request a certain range of bytes, but there are some consequences:
it depends on the server to honor the request
you assume that data you're looking for is within desired range (however you can make another request using different range header to get next bytes - i.e. download first 300 bytes and get another 300 only if you can't find title within first result - 2 requests of 300 bytes are still much cheaper than whole document)
(edit) - to avoid situations when title tag splits between two ranged requests, make your ranges overlapped, see 'range_header_overlapped' function in my example code
import urllib
req = urllib.request.Request('http://www.python.org/')
req.headers['Range']='bytes=%s-%s' % (0, 300)
f = urllib.request.urlopen(req)
just to verify if server accepted our range:
content_range=f.headers.get('Content-Range')
print(content_range)
my code also solves cases when title tag is splitted between chunks.
#!/usr/bin/env python2
# -*- coding: utf-8 -*-
"""
Created on Tue May 30 04:21:26 2017
====================
#author: s
"""
import requests
from string import lower
from html.parser import HTMLParser
#proxies = { 'http': 'http://127.0.0.1:8080' }
urls = ['http://opencvexamples.blogspot.com/p/learning-opencv-functions-step-by-step.html',
'http://www.robindavid.fr/opencv-tutorial/chapter2-filters-and-arithmetic.html',
'http://blog.iank.org/playing-capitals-with-opencv-and-python.html',
'http://docs.opencv.org/3.2.0/df/d9d/tutorial_py_colorspaces.html',
'http://scikit-image.org/docs/dev/api/skimage.exposure.html',
'http://apprize.info/programming/opencv/8.html',
'http://opencvexamples.blogspot.com/2013/09/find-contour.html',
'http://docs.opencv.org/2.4/modules/imgproc/doc/geometric_transformations.html',
'https://github.com/ArunJayan/OpenCV-Python/blob/master/resize.py']
class TitleParser(HTMLParser):
def __init__(self):
HTMLParser.__init__(self)
self.match = False
self.title = ''
def handle_starttag(self, tag, attributes):
self.match = True if tag == 'title' else False
def handle_data(self, data):
if self.match:
self.title = data
self.match = False
def valid_content( url, proxies=None ):
valid = [ 'text/html; charset=utf-8',
'text/html',
'application/xhtml+xml',
'application/xhtml',
'application/xml',
'text/xml' ]
r = requests.head(url, proxies=proxies)
our_type = lower(r.headers.get('Content-Type'))
if not our_type in valid:
print('unknown content-type: {} at URL:{}'.format(our_type, url))
return False
return our_type in valid
def range_header_overlapped( chunksize, seg_num=0, overlap=50 ):
"""
generate overlapping ranges
(to solve cases when title tag splits between them)
seg_num: segment number we want, 0 based
overlap: number of overlaping bytes, defaults to 50
"""
start = chunksize * seg_num
end = chunksize * (seg_num + 1)
if seg_num:
overlap = overlap * seg_num
start -= overlap
end -= overlap
return {'Range': 'bytes={}-{}'.format( start, end )}
def get_title_from_url(url, proxies=None, chunksize=300, max_chunks=5):
if not valid_content(url, proxies=proxies):
return False
current_chunk = 0
myparser = TitleParser()
while current_chunk <= max_chunks:
headers = range_header_overlapped( chunksize, current_chunk )
headers['Accept-Encoding'] = 'deflate'
# quick fix, as my locally hosted Apache/2.4.25 kept raising
# ContentDecodingError when using "Content-Encoding: gzip"
# ContentDecodingError: ('Received response with content-encoding: gzip, but failed to decode it.',
# error('Error -3 while decompressing: incorrect header check',))
r = requests.get( url, headers=headers, proxies=proxies )
myparser.feed(r.content)
if myparser.title:
return myparser.title
current_chunk += 1
print('title tag not found within {} chunks ({}b each) at {}'.format(current_chunk-1, chunksize, url))
return False
I'm writing a script to DL the entire collection of BBC podcasts from various show hosts. My script uses BS4, Mechanize, and wget.
I would like to know how I can test if a request for a URL yields a response code of '404' form the server. I have wrote the below function:
def getResponseCode(br, url):
print("Opening: " + url)
try:
response = br.open(url)
print("Response code: " + str(response.code))
return True
except (mechanize.HTTPError, mechanize.URLError) as e:
if isinstance(e,mechanize.HTTPError):
print("Mechanize error: " + str(e.code))
else:
print("Mechanize error: " + str(e.reason.args))
return False
I pass into it my Browser() object and a URL string. It returns either True or False depending on whether the response is a '404' or '200' (well actually, Mechanize throws and Exception if it is anything other than a '200' hence the Exception handling).
In main() I am basically looping over this function passing in a number of URLs from a list of URLs that I have scraped with BS4. When the function returns True I proceed to download the MP3 with wget.
However. My problem is:
The URLs are direct path to the podcast MP3 files on the remote
server and I have noticed that when the URL is available,
br.open(<URL>) will hang. I suspect this is because Mechanize is
caching/downloading the actual data from the server. I do not want
this because I merely want to return True if the response code is
'200'. How can I not cache/DL and just test the response code?
I have tried using br.open_novisit(url, data=None) however the hang still persists...
I don't think there's any good way to get Mechanize to do what you want. The whole point of Mechanize is that it's trying to simulate a browser visiting a URL, and a browser visiting a URL downloads the page. If you don't want to do that, don't use an API designed for that.
On top of that, whatever API you're using, by sending a GET request for the URL, you're asking the server to send you the entire response. Why do that just to hang up on it as soon as possible? Use the HEAD request to ask the server whether it's available. (Sometimes servers won't HEAD things even when they should, so you'll have to fall back to GET. But cross that bridge if you come to it.)
For example:
req = urllib.request.Request(url, method='HEAD')
resp = urllib.request.urlopen(req)
return 200 <= resp.code < 300
But this raises a question:
When the function returns True I proceed to download the MP3 with wget.
Why? Why not just use wget in the first place? If the URL is gettable, it will get the URL; if not, it will give you an error—just as easily as Mechanize will. And that avoids hitting each URL twice.
For that matter, why try to script wget, instead of using the built-in support in the stdlib or a third-party module like requests?
If you're just looking for a way to parallelize things, that's easy to do in Python:
def is_good_url(url):
req = urllib.request.Request(url, method='HEAD')
resp = urllib.request.urlopen(req)
return url, 200 <= resp.code < 300
with futures.ThreadPoolExecutor(max_workers=8) as executor:
fs = [executor.submit(is_good_url, url) for url in urls]
results = (f.result() for f in futures.as_completed(fs))
good_urls = [url for (url, good) in results if good]
And to change this to actually download the valid URLs instead of just making a note of which ones are valid, just change the task function to something that fetches and saves the data from a GET instead of doing the HEAD thing. The ThreadPoolExecutor Example in the docs does almost exactly what you want.
I've been working on a script and I thought I would ask for help. I'm looking to search a series of websites, check if the site is valid. Then the next step would be to check for specific content on the site. If the site holds that content, place the URL in a list.
import urllib2
def getPage():
url="import urllib2
National=[]
Local=[]
Sports=[]
Culture=[]
def getPage():
url="http://readingeagle.com/section.aspx?id=2"
for i in range (0,100,1)
req = urllib2.Request(http://readingeagle.com/section.aspx?id=,i)
if "national" in response:
response = urllib2.urlopen(req)
return response.read()
for g in range (0,100,1)
if "national" in response:
National.append("http://readingeagle.com/section.aspx?id=,g"
# I would like to set-up an iteration to check the 'entryid from 1-100. If the term is found on the page, place the url in the list.
if __name__ == "__main__":
namesPage = getPage()
print (namesPage)
Here's my answer to the question of how to validate a given web site.
python check html valid
For checking the context of the page the tools consist of basic string methods, regex, or more sophisticated tools like lxml or beautifulsoup.
matchingSites = []
matchingSites.append(url) #Since you asked. :-p