I'm trying to login in https://accounts.coursera.org/ using twill for python
I tried this sheet of code
import twill
b = get_browser()
b.go("https://accounts.coursera.org/")
b.showforms()
twill doesn't detect the form in the page and showforms methods doesn't show anything !!
Is that an internal issue in twill package or I'm misssing something
import twill
import webbrowser
b = twill.get_browser()
b.go("https://accounts.coursera.org/")
page = b.result.get_page()
tmp_page = "tmp.html"
with file(tmp_page, "w") as f:
f.write(page)
webbrowser.open(tmp_page)
# b.showforms()
I get a page that says..
Please use a modern browser with JavaScript enabled to use Coursera.
So I suspect that twill doesn't include a javascript interpreter?
Related
I have a subscription to the site https://www.naturalgasintel.com/ for daily feeds of data that show up on their site directly as .txt files; their user login page being https://www.naturalgasintel.com/user/login/
For example a file for today's feed is given by the link https://naturalgasintel.com/ext/resources/Data-Feed/Daily-GPI/2019/01/20190104td.txt and shows up on the site like the picture below:
What I'd like to do is to log in using my user_email and user_password and scrape this data in the form of an Excel file.
When I use Twill to try and 'point' me to the data by first logging me into the site I use this code:
from email.mime.text import MIMEText
from subprocess import Popen, PIPE
import twill
from twill.commands import *
year= NOW[0:4]
month=NOW[5:7]
day=NOW[8:10]
date=(year+month+day)
path = "https://naturalgasintel.com/ext/resources/Data-Feed/Daily-GPI/"
end = "td.txt"
go("http://www.naturalgasintel.com/user/login")
fv("2", "user[email]", user_email)
fv("2", "user[password]", user_password)
fv("2", "commit", "Login")
datafilelocation = path + year + "/" + month + "/" + date + end
go(datafilelocation)
However, logging in from the user login page sends me to this referrer link when I go to the data's location.
https://www.naturalgasintel.com/user/login?referer=%2Fext%2Fresources%2FData-Feed%2FDaily-GPI%2F2019%2F01%2F20190104td.txt
Rather than:
https://naturalgasintel.com/ext/resources/Data-Feed/Daily-GPI/2019/01/20190104td.txt
I've tried using modules like requests as well to log in from the site and then access this data but whatever method I use sends me to the HTML source rather than the .txt data location itself.
I've posted my complete walk-through with the Python 2.7 module Twill which I attached a bounty to here:
Using Twill to grab .txt from login page Python
What would the best solution to being able to access these password protected files be?
If you have a compatible version of FireFox for this, then get the plugin javascript 0.0.1 by Chee and add the following to run on the page:
document.getElementById('user_email').value = "E-What";
document.getElementById('user_password').value = " ABC Password ";
Change the email and password as you like. It will load the page, then after that it will put in your username and password.
There are other ways to do this all by yourself with your own stand-alone process. You do not have to download other people's programs and try to learn them (beyond this little thing) if you change it this way.
I would have up voted this question.
I'm trying to automate the process of creating an account for something, lets call it X, but I cant figure out what to do.
I saw this code somewhere,
import urllib
import urllib2
import webbrowser
data = urllib.urlencode({'q': 'Python'})
url = 'http://duckduckgo.com/html/'
full_url = url + '?' + data
response = urllib2.urlopen(full_url)
with open("results.html", "w") as f:
f.write(response.read())
webbrowser.open("results.html")
But I cant figure out how to modify it for my use.
I would highly recommend utilizing Selenium+Webdriver for this, since your question appears UI and browser-based. You can install Selenium via 'pip install selenium' in most cases. Here are a couple of good references to get started.
- http://selenium-python.readthedocs.io/
- https://pypi.python.org/pypi/selenium
Also, if this process needs to drive the browser headlessly, look into including PhantomJS (via GhostDriver), which can be downloaded from the phantomjs.org website.
I'm building a Django app and I'm using Spynner for web crawling. I have this problem and I hope someone can help me.
I have this function in the module "crawler.py":
import spynner
def crawling_js(url)
br = spynner.Browser()
br.load(url)
text_page = br.html
br.close (*)
return text_page
(*) I tried with br.close() too
in another module (eg: "import.py") I call the function in this way:
from crawler import crawling_js
l_url = ["https://www.google.com/", "https://www.tripadvisor.com/", ...]
for url in l_url:
mytextpage = crawling_js(url)
.. parse mytextpage....
when I pass the first url in to the function all is correct when I pass the second "url" python crash. Python crash in this line:br.load(url). Someone can help me? Thanks a lot
I have:
Django 1.3
Python 2.7
Spynner 1.1.0
PyQt4 4.9.1
Why you need to instantiate br = spynner.Browser() and close it every time you call crawling_js(). In a loop this will utilize a lot of resources which I think is the reason why it crashes. let's think of it like this, br is a browser instance. Therefore, you can make it browse any number of websites without the need to close it and open it again. Adjust your code this way:
import spynner
br = spynner.Browser() #you open it only once.
def crawling_js(url):
br.load(url)
text_page = br._get_html() #_get_html() to make sure you get the updated html
return text_page
then if you insist to close br later you simply do:
from crawler import crawling_js , br
l_url = ["https://www.google.com/", "https://www.tripadvisor.com/", ...]
for url in l_url:
mytextpage = crawling_js(url)
.. parse mytextpage....
br.close()
I am writing a python script for scraping a webpage. I have created a webkit webview object and used the open method for loading the url. But I want to load the url through a proxy.
How can i done this ? How to integrate webkit with proxy? which webkit class support proxy?
try below code snippets. (reference from url)
import gtk, webkit
import ctypes
libgobject = ctypes.CDLL('/usr/lib/libgobject-2.0.so.0')
libwebkit = ctypes.CDLL('/usr/lib/libsoup-2.4.so.1')
libsoup = ctypes.CDLL('/usr/lib/libsoup-2.4.so.1')
libwebkit = ctypes.CDLL('/usr/lib/libwebkit-1.0.so')
proxy_uri = libsoup.soup_uri_new('http://127.0.0.1:8000') # set your proxy url
session = libwebkit.webkit_get_default_session()
libgobject.g_object_set(session, "proxy-uri", proxy_uri, None)
w = gtk.Window()
s = gtk.ScrolledWindow()
v = webkit.WebView()
s.add(v)
w.add(s)
w.show_all()
v.open('http://www.google.com')
Hope, it could help you.
You can use QApplicationProxy if you're on pyqt or this snippet if you're using pygi:
from gi.repository import WebKit
from gi.repository import Soup
proxy_uri = Soup.URI.new("http://127.0.0.1:8080")
session = WebKit.get_default_session().set_property("proxy-uri")
session.set_property("proxy-uri",proxy_uri)
References:
PyGI
PyQt
How about a solution that's already made?
PyPhantomJS is a minimalistic, headless, WebKit-based, JavaScript-driven tool. It is written in PyQt4 and Python. It runs on Linux, Windows, and Mac OS X.
It gives you access to a full headless WebKit browser, controllable via scripts written in JavaScript, with the ability to do various things, amongst which is screen scraping and proxy support. It uses the command line.
You can see the API here.
* When I say screen scraping, I mean you can either scrape page content, or even save page renders to a file. There's even a screen scraping JS library already written here.
I'm currently trying to get a grasp on pycurl. I'm attempting to login to a website. After logging into the site it should redirect to the main page. However when trying this script it just gets returned to the login page. What might I be doing wrong?
import pycurl
import urllib
import StringIO
pf = {'username' : 'user', 'password' : 'pass' }
fields = urllib.urlencode(pf)
pageContents = StringIO.StringIO()
p = pycurl.Curl()
p.setopt(pycurl.FOLLOWLOCATION, 1)
p.setopt(pycurl.COOKIEFILE, './cookie_test.txt')
p.setopt(pycurl.COOKIEJAR, './cookie_test.txt')
p.setopt(pycurl.POST, 1)
p.setopt(pycurl.POSTFIELDS, fields)
p.setopt(pycurl.WRITEFUNCTION, pageContents.write)
p.setopt(pycurl.URL, 'http://localhost')
p.perform()
pageContents.seek(0)
print pageContents.readlines()
EDIT: As pointed out by Peter the URL should point to a login URL but the site I'm trying to get this to work for fails to show me what URL this would be. The form's action just points to the home page ( /index.html )
As you're troubleshooting this problem, I suggest getting a browser plugin like FireBug or LiveHTTPHeaders (I suggest Firefox plugins, but there are similar plugins for other browsers as well). Then you can exercise a request to the site and see what action (URL), method, and form parameters are being passed to the target server. This will likely help elucidate the crux of the problem.
If that's no help, you may consider using a different tool for your mechanization. I've used ClientForm and BeautifulSoup to perform similar operations. Based on what I've read in the pycURL docs and your code above, ClientForm might be a better tool to use. ClientForm will parse your HTML page, locate the forms on it (including login forms), and construct the appropriate request for you based on the answers you supply to the form. You could even use ClientForm with pycURL... but at least ClientForm will provide you with the appropriate action to which to POST, and construct all of the appropriate parameters.
Be aware, though, that if there is JavaScript handling any necessary part of the login form, even ClientForm can't help you there. You will need something that interprets the JavaScript to effectively automate the login. In that case, I've used SeleniumRC to control a browser (and I let the browser handle the JavaScript).
One of the golden rule, you need to 'brake the ice', have debugging enabled when trying to solve pycurl example:
Note: don't forget to use p.close() after p.perform()
def test(debug_type, debug_msg):
if len(debug_msg) < 300:
print "debug(%d): %s" % (debug_type, debug_msg.strip())
p.setopt(pycurl.VERBOSE, True)
p.setopt(pycurl.DEBUGFUNCTION, test)
Now you can see how your code is breathing, because you have debugging enabled
import pycurl
import urllib
import StringIO
def test(debug_type, debug_msg):
if len(debug_msg) < 300:
print "debug(%d): %s" % (debug_type, debug_msg.strip())
pf = {'username' : 'user', 'password' : 'pass' }
fields = urllib.urlencode(pf)
pageContents = StringIO.StringIO()
p = pycurl.Curl()
p.setopt(pycurl.FOLLOWLOCATION, 1)
p.setopt(pycurl.COOKIEFILE, './cookie_test.txt')
p.setopt(pycurl.COOKIEJAR, './cookie_test.txt')
p.setopt(pycurl.POST, 1)
p.setopt(pycurl.POSTFIELDS, fields)
p.setopt(pycurl.WRITEFUNCTION, pageContents.write)
p.setopt(pycurl.VERBOSE, True)
p.setopt(pycurl.DEBUGFUNCTION, test)
p.setopt(pycurl.URL, 'http://localhost')
p.perform()
p.close() # This is mandatory.
pageContents.seek(0)
print pageContents.readlines()