urllib2

I'm trying to get a queried-excel file from a site. When I enter the direct link, it will lead to a login page and once I've entered my username and password, it will proceed to download the excel file automatically. I am trying to avoid installing additional module that's not part of the standard python (This script will be running on a "standardize machine" and it won't work if the module is not installed) I've tried the following but I see a "page login" information in the excel file itself :-| import urllib url = "myLink_queriedResult/result.xls" urllib.urlretrieve(url,"C:\\test.xls") SO..

I'm calling a list of urls from the same domain and returning a snip of their html for a few thousand domains but am getting this error about 1,000 rows or so in. Is there anything I can do to avoid this error? Does it make sense to create a wait step after every row? every few hundred rows? Is there a better way to get around this? File "/Users.../ap.py", line 144, in <module> simpleProg() File "/Users.../ap.py", line 21, in simpleProg() File "/Users.../ap.py", line 57, in first_step() File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 126, in urlopen

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I am trying to write my first Python script, and with lots of Googling, I think that I am just about done. However, I will need some help getting myself across the finish line. I need to write a script that logs onto a cookie-enabled site, scrape a bunch of links, and then spawn a few processes to download the files. I have the program running in single-threaded, so I know that the code works. But, when I tried to create a pool of download workers, I ran into a wall. #manager.py import Fetch # the module name where worker lives from multiprocessing import pool def FetchReports(links,Username

The following python curl call has the following successful results: >>> import subprocess >>> args = [ 'curl', '-H', 'X-Requested-With: Demo', 'https://username:password@qualysapi.qualys.com/qps/rest/3.0/count/was/webapp' ] >>> xml_output = subprocess.check_output(args).decode('utf-8') % Total % Received % Xferd Average Speed Time Time Time Current Dload Upload Total Spent Left Speed 138 276 0 276 0 0 190 0 --:--:-- 0:00:01 --:--:-- 315 >>> xml_output u'<?xml version="1.0" encoding="UTF-8"?>\n<ServiceResponse xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:noNamespaceSchemaLocation=

I am wishing to extract the content of pdf files available online using PDFMiner . My code is based on the one available in the documentation used to extract the content of PDF files on the hard disk: # Open a PDF file. fp = open('mypdf.pdf', 'rb') # Create a PDF parser object associated with the file object. parser = PDFParser(fp) # Create a PDF document object that stores the document structure. document = PDFDocument(parser) That works quite well with some small changes. Now, I have tried urllib2.openurl for online PDFs but that doesn't work. I get an error message : coercing to Unicode:

I'm using python 2.7 and I'd like to get the contents of a webpage that requires sslv3. Currently when I try to access the page I get the error SSL23_GET_SERVER_HELLO, and some searching on the web lead me to the following solution which fixes things in Python 3 urllib.request.install_opener(urllib.request.build_opener(urllib.request.HTTPSHandler(context=ssl.SSLContext(ssl.PROTOCOL_TLSv1)))) How can I get the same effect in python 2.7, as I can't seem to find the equivalent of the context argument for the HTTPSHandler class. I realize this response is a few years too late, but I also ran into

I would like to get the estimated results number from google for a keyword. Im using Python3.3 and try to accomplish this task with BeautifulSoup and urllib.request. This is my simple code so far def numResults(): try: page_google = '''http://www.google.de/#output=search&sclient=psy-ab&q=pokerbonus&oq=pokerbonus&gs_l=hp.3..0i10l2j0i10i30l2.16503.18949.0.20819.10.9.0.1.1.0.413.2110.2-6j1j1.8.0....0...1c.1.19.psy-ab.FEBvxrgi0KU&pbx=1&bav=on.2,or.r_qf.&bvm=bv.48705608,d.Yms&''' req_google = Request(page_google) req_google.add_header('User Agent', 'Mozilla/5.0 (Windows NT 6.1; WOW64; rv:15.0)

I am running a script using python that uses urllib2 to grab data from a weather api and display it on screen. I have had the problem that when I query the server, I get a "no address associated with hostname" error. I can view the output of the api with a web browser, and I can download the file with wget, but I have to force IPv4 to get it to work. Is it possible to force IPv4 in urllib2 when using urllib2.urlopen? Not directly, no. So, what can you do? One possibility is to explicitly resolve the hostname to IPv4 yourself, and then use the IPv4 address instead of the name as the host. For