I am setting up code to check the reputation of any URL E.g. http://go.mobisla.com/ on Website "https://www.virustotal.com/gui/home/url"
First, the very basic thing I am doing is to extract all the Website contents using BeautifulSoup but seems the information I am looking for is in shadow-root(open) -- div.detections and span.individual-detection.
Example Copied Element from Webpage results:
No engines detected this URL
I am new to Python, wondering if you can share the best way to extract the information
Tried requests.get() function but it doesn't give the required information
import requests
import os,sys
from bs4 import BeautifulSoup
import pandas as pd
url_check = "deloplen.com:443"
url = "https://www.virustotal.com/gui/home/url"
req = requests.get(url + url_str)
html = req.text
soup = BeautifulSoup(html, 'html.parser')
print(soup.prettify())
Expect to see "2 engines detected this URL" along with Detection Example: Dr. Web Malicious
If you use their website, it'll only return a loading screen for VirusTotal, as this isn't the proper way.
What Shows Up:
Instead, what you're supposed to do is use their public API to make requests. However, you'll have to make an account to obtain a Public API Key.
You can use this code which is able to retrieve JSON info about the link. However, you'll have to fill in the API KEY with yours.
import requests, json
user_api_key = "<api key>"
resource = "deloplen.com:443"
# feel free to remove this, just makes it look nicer
def pp_json(json_thing, sort=True, indents=4):
if type(json_thing) is str:
print(json.dumps(json.loads(json_thing), sort_keys=sort, indent=indents))
else:
print(json.dumps(json_thing, sort_keys=sort, indent=indents))
return None
response = requests.get("https://www.virustotal.com/vtapi/v2/url/report?apikey=" + user_api_key + "&resource=" + resource)
json_response = response.json()
pretty_json = pp_json(json_response)
print(pretty_json)
If you want to learn more about the API, you can use their documentation.
The code below queries the Wikipedia API for pages in the "Physics" category and converts the response into a Python dictionary.
import ast
import requests
url = "https://en.wikipedia.org/w/api.php?action=query&list=categorymembers&cmtitle=Category:Physics&cmlimit=500&cmcontinue="
response = requests.get(url)
text = response.text
dict = ast.literal_eval(sourceCode)
Here is one of the results returned by the Wikipedia API:
{
"pageid": 50724262,
"ns": 0,
"title": "Blasius\u2013Chaplygin formula"
},
The Wikipedia page that "Blasius\u2013Chaplygin formula" corresponds to is https://en.wikipedia.org/wiki/Blasius–Chaplygin_formula.
I want to use the "title" to download pages from Wikipedia. I've replaced all spaces with underscores. But it's failing. I'm doing:
import requests
url = "https://en.wikipedia.org/wiki/Blasius\u2013Chaplygin_formula"
response = requests.get(url)
This gives me:
requests.exceptions.HTTPError: 404 Client Error:
Not Found for url: https://en.wikipedia.org/wiki/Blasius%5Cu2013Chaplygin_formula
How do I change the title Blasius\u2013Chaplygin formula into a URL that can be successfully called by requests?
When I tried to insert the Wikipedia link into this question on Stack Overflow, Stack Overflow automatically converted it to https://en.wikipedia.org/wiki/Blasius%E2%80%93Chaplygin_formula.
When I did:
import requests
url = "https://en.wikipedia.org/wiki/Blasius%E2%80%93Chaplygin_formula"
response = requests.get(url)
it was successful, so I want a library that will do a conversion like this that I can use in Python.
That "\u2013" is a unicode character. It gets automatically turned into an en-dash by python, but you can't put en-dashes in wikipedia links, so you have to url encode it, which is what stackoverflow did for you earlier.
You can do it yourself by using something like this:
import requests
import urllib.parse
url = "Blasius\u2013Chaplygin_formula"
response = requests.get("https://en.wikipedia.org/wiki/" + urllib.parse.quote(url))
How to urlencode a querystring in Python?
To make your life easier you can always use some existing wrapper around Wikipedia API such as Wikipedia-API.
import wikipediaapi
api = wikipediaapi.Wikipedia('en')
# it will shield you from URL encoding problems
p = api.page('Blasius\u2013Chaplygin formula')
print(p.summary)
# and it can make your code shorter
physics = api.page('Category:Physics')
for p in physics.categorymembers.values():
print(f'[{p.title}]\t{p.summary}')
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
Hi I am building a blogging website in django 1.8 with python 3. In the blog users will write blogs and sometimes add external links.
I want to crawl all the pages in this blog website and test every external link provided by the users is valid or not.
How can i do this? Should i use something like python scrapy?
import urllib2
import fnmatch
def site_checker(url):
url_chk = url.split('/')
if fnmatch.fnmatch(url_chk[0], 'http*'):
url = url
else:
url = 'http://%s' %(url)
print url
try:
response = urllib2.urlopen(url).read()
if response:
print 'site is legit'
except Exception:
print "not a legit site yo!"
site_checker('google') ## not a complete url
site_checker('http://google.com') ## this works
Hopefully this works. Urllib will read the html of the site and if its not empty. It's a legit site. Else it's not a site. Also I added a url check to add http:// if its not there.
As of now I have created a basic program in python 2.7 using urllib2 and re that gathers the html code of a website and prints it out for you as well as indexing a keyword. I would like to create a much more complex and dynamic program which could gather data from websites such as sports or stock statistics and aggregate them into lists which could then be used in analysis in something such as an excel document etc. I'm not asking for someone to literally write the code. I simply need help understanding more of how I should approach the code: whether I require extra libraries, etc. Here is the current code. It is very simplistic as of now.:
import urllib2
import re
y = 0
while(y == 0):
x = str(raw_input("[[[Enter URL]]]"))
keyword = str(raw_input("[[[Enter Keyword]]]"))
wait = 0
try:
req = urllib2.Request(x)
response = urllib2.urlopen(req)
page_content = response.read()
idall = [m.start() for m in re.finditer(keyword,page_content)]
wait = raw_input("")
print(idall)
wait = raw_input("")
print(page_content)
except urllib2.HTTPError as e:
print e.reason
You can use requests to deal with interaction with website. Here is link for it. http://docs.python-requests.org/en/latest/
And then you can use beautifulsoup to handle the html content. Here is link for it.http://www.crummy.com/software/BeautifulSoup/bs4/doc/index.zh.html
They're more ease of use than urllib2 and re.
Hope it helps.