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  • How to use urllib2 when users only have a API token?

    - by jorrebor
    how would i tranfoms this curl command: curl -v -u 82xxxxxxxxxxxx63e6:api_token -X GET https://www.toggl.com/api/v6/time_entries.json into urlib2? I found this tutorial: http://www.voidspace.org.uk/python/articles/authentication.shtml but they use a password and username. I can only use an API token. Thank you. see also this question: Urllib2 raises 403 error while the same request in curl works fine

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  • Difference between URLLIB2 call in IDLE and from Django?

    - by danspants
    The following piece of code works as expected when running in a local install of django apache 2.2 fx = urllib2.Request(f); fx.add_header('User-Agent','Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US) AppleWebKit/525.19 (KHTML, like Gecko) Chrome/1.0.154.36 Safari/525.19'); url_opened = urllib2.urlopen(fx); However when I enter that code into IDLE on the same machine I get the following error: url_opened = urllib2.urlopen(fx); File "C:\Python25\lib\urllib2.py", line 124, in urlopen return _opener.open(url, data) File "C:\Python25\lib\urllib2.py", line 387, in open response = meth(req, response) File "C:\Python25\lib\urllib2.py", line 498, in http_response 'http', request, response, code, msg, hdrs) File "C:\Python25\lib\urllib2.py", line 425, in error return self._call_chain(*args) File "C:\Python25\lib\urllib2.py", line 360, in _call_chain result = func(*args) File "C:\Python25\lib\urllib2.py", line 506, in http_error_default raise HTTPError(req.get_full_url(), code, msg, hdrs, fp) HTTPError: HTTP Error 407: Proxy Authentication Required Any ideas?

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  • Bitbucket API authentication with Python's HTTPBasicAuthHandler

    - by jbochi
    I'm trying to get the list of issues on a private repository using bitbucket's API. I have confirmed that HTTP Basic authentication works with hurl, but I am unable to authenticate in Python. Adapting the code from this tutorial, I have written the following script. import cookielib import urllib2 class API(): api_url = 'http://api.bitbucket.org/1.0/' def __init__(self, username, password): self._opener = self._create_opener(username, password) def _create_opener(self, username, password): cj = cookielib.LWPCookieJar() cookie_handler = urllib2.HTTPCookieProcessor(cj) password_manager = urllib2.HTTPPasswordMgrWithDefaultRealm() password_manager.add_password(None, self.api_url, username, password) auth_handler = urllib2.HTTPBasicAuthHandler(password_manager) opener = urllib2.build_opener(cookie_handler, auth_handler) return opener def get_issues(self, username, repository): query_url = self.api_url + 'repositories/%s/%s/issues/' % (username, repository) try: handler = self._opener.open(query_url) except urllib2.HTTPError, e: print e.headers raise e return handler.read() api = API(username='my_username', password='XXXXXXXX') api.get_issues('my_username', 'my_repository') results in: >>> Server: nginx/0.7.62 Date: Mon, 19 Apr 2010 16:15:06 GMT Content-Type: text/plain Connection: close Vary: Authorization,Cookie Content-Length: 9 Traceback (most recent call last): File "C:/USERS/personal/bitbucket-burndown/bitbucket-api.py", line 29, in <module> print api.get_issues('my_username', 'my_repository') File "C:/USERS/personal/bitbucket-burndown/bitbucket-api.py", line 25, in get_issues raise e HTTPError: HTTP Error 401: UNAUTHORIZED api.get_issues('jespern', 'bitbucket') works like a charm. What's wrong with my code?

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  • Python 2.6 -> Python 3 (ProxyHandler)

    - by blah
    Hallo, I wrote a script that works with a proxy (py2.6x): proxy_support = urllib2.ProxyHandler({'http' : 'http://127.0.0.1:80'}) But in py3.11x there is no urllib2 just a urllib... and that doesnt support the ProxyHandler How can I use a proxy with urllib? Isnt Python 3 newer then Python 2? Why did they remove urllib2 in a newer version?

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  • Storing cookielib cookies in a database

    - by Mridang Agarwalla
    Hi, I'm using the cookielib module to handle HTTP cookies when using the urllib2 module in Python 2.6 in a way similar to this snippet: import cookielib, urllib2 cj = cookielib.CookieJar() opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(cj)) r = opener.open("http://example.com/") I'd like to store the cookies in a database. I don't know whats better - serialize the CookieJar object and store it or extract the cookies from the CookieJar and store that. I don't know which one's better or how to implement either of them. I should be also be able to recreate the CookieJar object. Could someone help me out with the above? Thanks in advance.

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  • Python: HTTP Post a large file with streaming

    - by Daniel Von Fange
    I'm uploading potentially large files to a web server. Currently I'm doing this: import urllib2 f = open('somelargefile.zip','rb') request = urllib2.Request(url,f.read()) request.add_header("Content-Type", "application/zip") response = urllib2.urlopen(request) However, this reads the entire file's contents into memory before posting it. How can I have it stream the file to the server?

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  • Twitter API with urllib2 in python

    - by Dirk Nachbar
    I want to use the Twitter API in Python to lookup user ids from name using the lookup method. I have done similar requests simply using response = urllib2.urlopen('http://search.twitter.com...') but for this one I need authentication. I don't think I can do it through the Google python twitter API because it doesn't have the lookup method. Any ideas how can I can auth with urllib2??

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  • deploying a war to tomcat using python

    - by Decado
    Hi, I'm trying to deploy a war to a Apache Tomcat server (Build 6.0.24) using python (2.4.2) as part of a build process. I'm using the following code import urllib2 import base64 war_file_contents = open('war_file.war','rb').read() username='some_user' password='some_pwd' base64string = base64.encodestring('%s:%s' % (username, password))[:-1] authheader = "Basic %s" % base64string opener = urllib2.build_opener(urllib2.HTTPHandler) request = urllib2.Request('http://158.155.40.110:8080/manager/deploy?path=war_file', data=war_file_contents) request.add_header('Content-Type', 'application/octet-stream') request.add_header("Authorization", authheader) request.get_method = lambda: 'PUT' url = opener.open(request) the url.code is 200, and the url.msg is "OK". However the web archive doesn't appear on the manager list applications page. Thanks.

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  • Why can't I download a whole image file with urllib2.urlopen()

    - by John Gann
    When I run the following code, it only seems to be downloading the first little bit of the file and then exiting. Occassionally, I will get a 10054 error, but usually it just exits without getting the whole file. My internet connection is crappy wireless, and I often get broken downloads on larger files in firefox, but my browser has no problem getting a 200k image file. I'm new to python, and programming in general, so I'm wondering what nuance I'm missing. import urllib2 xkcdpic=urllib2.urlopen("http://imgs.xkcd.com/comics/literally.png") xkcdpicfile=open("C:\\Documents and Settings\\John Gann\\Desktop\\xkcd.png","w") while 1: chunk=xkcdpic.read(4028) if chunk: print chunk xkcdpicfile.write(chunk) else: break

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  • Urllib's urlopen breaking on some sites (e.g. StackApps api)

    - by Edan Maor
    I'm using urllib2's urlopen function to try and get a JSON result from the StackOverflow api. The code I'm using: >>> import urllib2 >>> conn = urllib2.urlopen("http://api.stackoverflow.com/0.8/users/") >>> conn.readline() The result I'm getting: '\x1f\x8b\x08\x00\x00\x00\x00\x00\x04\x00\xed\xbd\x07`\x1cI\x96%&/m\xca{\x7fJ\... I'm fairly new to urllib, but this doesn't seem like the result I should be getting. I've tried it in other places and I get what I expect (the same as visiting the address with a browser gives me: a JSON object). Using urlopen on other sites (e.g. "http://google.com") works fine, and gives me actual html. I've also tried using urllib and it gives the same result. I'm pretty stuck, not even knowing where to look to solve this problem. Any ideas?

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  • Urllib's urlopen broken on some sites (e.g. StackApps api)

    - by Edan Maor
    I'm using urllib2's urlopen function to try and get a JSON result from the StackOverflow api. The code I'm using: >>> import urllib2 >>> conn = urllib2.urlopen("http://api.stackoverflow.com/0.8/users/") >>> conn.readline() The result I'm getting: '\x1f\x8b\x08\x00\x00\x00\x00\x00\x04\x00\xed\xbd\x07`\x1cI\x96%&/m\xca{\x7fJ\... I'm fairly new to urllib, but this doesn't seem like the result I should be getting. I've tried it in other places and I get what I expect (the same as visiting the address with a browser gives me: a JSON object). Using urlopen on other sites (e.g. "http://google.com") works fine, and gives me actual html. I've also tried using urllib and it gives the same result. I'm pretty stuck, not even knowing where to look to solve this problem. Any ideas?

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  • Urllib's urlopen breaking on some sites (e.g. StackApps api): returns garbage results

    - by Edan Maor
    I'm using urllib2's urlopen function to try and get a JSON result from the StackOverflow api. The code I'm using: >>> import urllib2 >>> conn = urllib2.urlopen("http://api.stackoverflow.com/0.8/users/") >>> conn.readline() The result I'm getting: '\x1f\x8b\x08\x00\x00\x00\x00\x00\x04\x00\xed\xbd\x07`\x1cI\x96%&/m\xca{\x7fJ\... I'm fairly new to urllib, but this doesn't seem like the result I should be getting. I've tried it in other places and I get what I expect (the same as visiting the address with a browser gives me: a JSON object). Using urlopen on other sites (e.g. "http://google.com") works fine, and gives me actual html. I've also tried using urllib and it gives the same result. I'm pretty stuck, not even knowing where to look to solve this problem. Any ideas?

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  • Urllib's urlopen broken on some sites (StackApps api)

    - by Edan Maor
    I'm using urllib2's urlopen function to try and get a JSON result from the StackOverflow api. The code I'm using: >>> import urllib2 >>> conn = urllib2.urlopen("http://api.stackoverflow.com/0.8/users/") >>> conn.readline() The result I'm getting: '\x1f\x8b\x08\x00\x00\x00\x00\x00\x04\x00\xed\xbd\x07`\x1cI\x96%&/m\xca{\x7fJ\... I'm fairly new to urllib, but this doesn't seem like the result I should be getting. I've tried it in other places and I get what I expect (the same as visiting the address with a browser gives me: a JSON object). Using urlopen on other sites (e.g. "http://google.com") works fine, and gives me actual html. I've also tried using urllib and it gives the same result. I'm pretty stuck, not even knowing where to look to solve this problem. Any ideas?

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  • Downloading a web page and all of its resource files in Python

    - by Mark
    I want to be able to download a page and all of its associated resources (images, style sheets, script files, etc) using Python. I am (somewhat) familiar with urllib2 and know how to download individual urls, but before I go and start hacking at BeautifulSoup + urllib2 I wanted to be sure that there wasn't already a Python equivalent to "wget --page-requisites http://www.google.com". Specifically I am interested in gathering statistical information about how long it takes to download an entire web page, including all resources. Thanks Mark

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  • Python interface to PayPal - urllib.urlencode non-ASCII characters failing

    - by krys
    I am trying to implement PayPal IPN functionality. The basic protocol is as such: The client is redirected from my site to PayPal's site to complete payment. He logs into his account, authorizes payment. PayPal calls a page on my server passing in details as POST. Details include a person's name, address, and payment info etc. I need to call a URL on PayPal's site internally from my processing page passing back all the params that were passed in abovem and an additional one called 'cmd' with a value of '_notify-validate'. When I try to urllib.urlencode the params which PayPal has sent to me, I get a: While calling send_response_to_paypal. Traceback (most recent call last): File "<snip>/account/paypal/views.py", line 108, in process_paypal_ipn verify_result = send_response_to_paypal(params) File "<snip>/account/paypal/views.py", line 41, in send_response_to_paypal params = urllib.urlencode(params) File "/usr/local/lib/python2.6/urllib.py", line 1261, in urlencode v = quote_plus(str(v)) UnicodeEncodeError: 'ascii' codec can't encode character u'\ufffd' in position 9: ordinal not in range(128) I understand that urlencode does ASCII encoding, and in certain cases, a user's contact info can contain non-ASCII characters. This is understandable. My question is, how do I encode non-ASCII characters for POSTing to a URL using urllib2.urlopen(req) (or other method) Details: I read the params in PayPal's original request as follows (the GET is for testing): def read_ipn_params(request): if request.POST: params= request.POST.copy() if "ipn_auth" in request.GET: params["ipn_auth"]=request.GET["ipn_auth"] return params else: return request.GET.copy() The code I use for sending back the request to PayPal from the processing page is: def send_response_to_paypal(params): params['cmd']='_notify-validate' params = urllib.urlencode(params) req = urllib2.Request(PAYPAL_API_WEBSITE, params) req.add_header("Content-type", "application/x-www-form-urlencoded") response = urllib2.urlopen(req) status = response.read() if not status == "VERIFIED": logging.warn("PayPal cannot verify IPN responses: " + status) return False return True Obviously, the problem only arises if someone's name or address or other field used for the PayPal payment does not fall into the ASCII range.

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  • use proxy in python to fetch a webpage

    - by carmao
    I am trying to write a function in Python to use a public anonymous proxy and fetch a webpage, but I got a rather strange error. The code (I have Python 2.4): import urllib2 def get_source_html_proxy(url, pip, timeout): # timeout in seconds (maximum number of seconds willing for the code to wait in # case there is a proxy that is not working, then it gives up) proxy_handler = urllib2.ProxyHandler({'http': pip}) opener = urllib2.build_opener(proxy_handler) opener.addheaders = [('User-agent', 'Mozilla/5.0')] urllib2.install_opener(opener) req=urllib2.Request(url) sock=urllib2.urlopen(req) timp=0 # a counter that is going to measure the time until the result (webpage) is # returned while 1: data = sock.read(1024) timp=timp+1 if len(data) < 1024: break timpLimita=50000000 * timeout if timp==timpLimita: # 5 millions is about 1 second break if timp==timpLimita: print IPul + ": Connection is working, but the webpage is fetched in more than 50 seconds. This proxy returns the following IP: " + str(data) return str(data) else: print "This proxy " + IPul + "= good proxy. " + "It returns the following IP: " + str(data) return str(data) # Now, I call the function to test it for one single proxy (IP:port) that does not support user and password (a public high anonymity proxy) #(I put a proxy that I know is working - slow, but is working) rez=get_source_html_proxy("http://www.whatismyip.com/automation/n09230945.asp", "93.84.221.248:3128", 50) print rez The error: Traceback (most recent call last): File "./public_html/cgi-bin/teste5.py", line 43, in ? rez=get_source_html_proxy("http://www.whatismyip.com/automation/n09230945.asp", "93.84.221.248:3128", 50) File "./public_html/cgi-bin/teste5.py", line 18, in get_source_html_proxy sock=urllib2.urlopen(req) File "/usr/lib64/python2.4/urllib2.py", line 130, in urlopen return _opener.open(url, data) File "/usr/lib64/python2.4/urllib2.py", line 358, in open response = self._open(req, data) File "/usr/lib64/python2.4/urllib2.py", line 376, in _open '_open', req) File "/usr/lib64/python2.4/urllib2.py", line 337, in _call_chain result = func(*args) File "/usr/lib64/python2.4/urllib2.py", line 573, in lambda r, proxy=url, type=type, meth=self.proxy_open: \ File "/usr/lib64/python2.4/urllib2.py", line 580, in proxy_open if '@' in host: TypeError: iterable argument required I do not know why the character "@" is an issue (I have no such in my code. Should I have?) Thanks in advance for your valuable help.

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  • virtualenv does not copy standard modules like shutil and urllib2

    - by Barthelemy
    When I create a new virtualenv, virtualenv .virtualenvs/my_env, there is only a subset of the standard python modules copied/linked to the new virtualenv. For example, when I do ls -l in .virtualenvs/my_env/lib/python2.6, I see: ... ... os.py -> /usr/lib/python2.6/os.py ... os.pyc -> /usr/lib/python2.6/os.pyc but modules like shutil and urllib2 are not copied even if they are in /usr/lib/python2.6/shutil.py. I am using Ubuntu 9.10. Is this the expected behavior? How can I install modules such as shutil in a virtualenv (I could not find these modules on pypi)?

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  • Python: Getting INVALID response from PayPal's Sandbox IPN, slowly going insane...

    - by thepeanut
    Hi All I am trying to implement a simple online payment system using PayPal, however I have tried everything I know and am still getting an INVALID response. I know it's nothing too simple, because I get a VERIFIED response when using the IPN simulator. I have tried putting the items into a dict first, I have tried fixing the encoding, and still nothing. PayPal says the reasons for an INVALID response could be: Sending wrong items or in wrong order (pretty sure it's not this) Sending to the wrong address (definitely not this) Encoding items incorrectly (I dont think it's this, set encoding to UTF-8 on both paypal and my script) The following is the snippet concerned: f = cgi.FieldStorage() newparams = 'cmd=_notify-validate' for key in f.keys(): val = f[key].value newparams += '&' + urlencode({key: val.encode('utf-8')}) req = urllib2.Request(PP_URL, newparams) req.add_header("Content-type", "application/x-www-form-urlencoded") http = urllib2.urlopen(req) ret = http.read() fi.write(ret + '\n') if ret == 'VERIFIED': #*do stuff* Can anyone suggest anything I can do to fix this?! Cheers Sam

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  • making urllib request in Python from the client side

    - by mridang
    Hi Guys, I've written a Python application that makes web requests using the urllib2 library after which it scrapes the data. I could deploy this as a web application which means all urllib2 requests go through my web-server. This leads to the danger of the server's IP being banned due to the high number of web requests for many users. The other option is to create an desktop application which I don't want to do. Is there any way I could deploy my application so that I can get my web-requests through the client side. One way was to use Jython to create an applet but I've read that Java applets can only make web-requests to the server it is deployed on and the only way to to circumvent this is to create a server side proxy which leads us back to the problem of the server's ip getting banned. This might sounds sound like and impossible situation and I'll probably end up creating a desktop application but I thought I'd ask if anyone knew of an alternate solution. Thanks.

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  • Manually extracting portions of strings contained in a list (parsing)

    - by user1652011
    I'm aware that there are modules that fully simplify this function, but saying that I am running from a base install of python (standard modules only), how would I extract the following: I have a list. This list is the contents, line by line, of a webpage. Here is a mock up list (unformatted) for informative purposes: <script> link = "/scripts/playlists/1/" + a.id + "/0-5417069212.asx"; <script> "<a href="/apps/audio/?feedId=11065"><span class="px13">Eastern Metro Area Fire</span>" From the above string, I need the following extracted. The feedId (11065), which is incidentally a.id in the code above., "/scripts/playlists/1/" and "/0-5417069212.asx". Remembering that each of these lines is just contents from objects in a list, how would I go about extracting that data? Here is the full list: contents = urllib2.urlopen("http://www.radioreference.com/apps/audio/?ctid=5586") Pseudo: from urllib2 import urlopen as getpage page_contents = getpage("http://www.radioreference.com/apps/audio/?ctid=5586") feedID = % in (page_contents.search() for "/apps/audio/?feedId=%") titleID = % in (page_contents.search() for "<span class="px13">%</span>") playlistID = % in (page_contents.search() for "link = "%" + a.id + "*.asx";") asxID = * in (page_contents.search() for "link = "*" + a.id + "%.asx";") streamURL = "http://www.radioreference.com/" + playlistID + feedID + asxID + ".asx" I plan to format it as such that streamURL should = : http://www.radioreference.com/scripts/playlists/1/11065/0-5417067072.asx

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  • download mbox files over https using python

    - by VenkatS
    I was trying to find the right module for downloading kernel patches from kernel.org site For example,to download the file at https://patchwork.kernel.org/patch/62948/mbox/ I understand urlgrabber has a problem with https on debian. urllib2 seems to have problem with this url as well (says getaddrinfo failed, even though there are no problems reaching other urls) Any help would be appreciated

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  • Python, implementing proxy support for a socket based application (not urllib2)

    - by Terry Felkrow
    Hey guys, I am little stumped: I have a simple messenger client program (pure python, sockets), and I wanted to add proxy support (http/s, socks), however I am a little confused on how to go about it. I am assuming that the connection on the socket level will be done to the proxy server, at which point the headers should contain a CONNECT + destination IP (of the chat server) and authentication, (if proxy requires so), however the rest is a little beyond me. How is the subsequent connection handled, specifically the reading/writing, etc... Are there any guides on proxy support implementation for socket based (tcp) programming in Python? Thank you

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  • How do I compile a Wikipedia lens and install?

    - by user49523
    I read a tutorial about how to compile and install a Wikipedia lens, but it didn't work. The tutorial sounds easy - i just copied and pasted to the file that was suppose to edit. I have tried some times and here are 2 edits edit 1: import logging import optparse import gettext from gettext import gettext as _ gettext.textdomain('wikipedia') from singlet.lens import SingleScopeLens, IconViewCategory, ListViewCategory from wikipedia import wikipediaconfig import urllib2 import simplejson class WikipediaLens(SingleScopeLens): wiki = "http://en.wikipedia.org" def wikipedia_query(self,search): try: search = search.replace(" ", "|") url = ("%s/w/api.php?action=opensearch&limit=25&format=json&search=%s" % (self.wiki, search)) results = simplejson.loads(urllib2.urlopen(url).read()) print "Searching Wikipedia" return results[1] except (IOError, KeyError, urllib2.URLError, urllib2.HTTPError, simplejson.JSONDecodeError): print "Error : Unable to search Wikipedia" return [] class Meta: name = 'Wikipedia' description = 'Wikipedia Lens' search_hint = 'Search Wikipedia' icon = 'wikipedia.svg' search_on_blank=True # TODO: Add your categories articles_category = ListViewCategory("Articles", "dialog-information-symbolic") def search(self, search, results): for article in self.wikipedia_query(search): results.append("%s/wiki/%s" % (self.wiki, article), "http://upload.wikimedia.org/wikipedia/commons/6/63/Wikipedia-logo.png", self.articles_category, "text/html", article, "Wikipedia Article", "%s/wiki/%s" % (self.wiki, article)) pass edit 2: import urllib2 import simplejson import logging import optparse import gettext from gettext import gettext as _ gettext.textdomain('wikipediaa') from singlet.lens import SingleScopeLens, IconViewCategory, ListViewCategory from wikipediaa import wikipediaaconfig class WikipediaaLens(SingleScopeLens): wiki = "http://en.wikipedia.org" def wikipedia_query(self,search): try: search = search.replace(" ", "|") url = ("%s/w/api.php?action=opensearch&limit=25&format=json&search=%s" % (self.wiki, search)) results = simplejson.loads(urllib2.urlopen(url).read()) print "Searching Wikipedia" return results[1] except (IOError, KeyError, urllib2.URLError, urllib2.HTTPError, simplejson.JSONDecodeError): print "Error : Unable to search Wikipedia" return [] def search(self, search, results): for article in self.wikipedia_query(search): results.append("%s/wiki/%s" % (self.wiki, article), "http://upload.wikimedia.org/wikipedia/commons/6/63/Wikipedia-logo.png", self.articles_category, "text/html", article, "Wikipedia Article", "%s/wiki/%s" % (self.wiki, article)) pass class Meta: name = 'Wikipedia' description = 'Wikipedia Lens' search_hint = 'Search Wikipedia' icon = 'wikipedia.svg' search_on_blank=True # TODO: Add your categories articles_category = ListViewCategory("Articles", "dialog-information-symbolic") def search(self, search, results): # TODO: Add your search results results.append('https://wiki.ubuntu.com/Unity/Lenses/Singlet', 'ubuntu-logo', self.example_category, "text/html", 'Learn More', 'Find out how to write your Unity Lens', 'https://wiki.ubuntu.com/Unity/Lenses/Singlet') pass so .. what can i change in the edit ? (if anybody give me the entire edit file edited i will appreciate)

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  • How to make a POST request with python-webkit?

    - by shakaran
    Hi, I new using python + webkit. I need make a POST request with webkit, but I dont know how to it. I use python-webkit because my app load a form on the GUI (for vote, comments and send more data) and I need post all these data with a POST request and load the html result send for the server to my GUI app with python-webkit. I have only this example with urllib: #!/usr/bin/python import urllib2, urllib import httplib server = 'server.somesite.com' data = {'name' : 'shakaran', 'password' : 'Only_I_know'} d = urllib.urlencode(data) headers = {"Content-type": "application/x-www-form- urlencoded", "Accept": "text/plain"} conn = httplib.HTTPConnection(server) conn.request("POST", "/login.php", d, headers) response = conn.getresponse() if response.status == 200: print response.status, response.reason print response.getheaders() data = response.read() print data conn.close() I need a simple example with webkit. I look in the documentation for Webkit.HTTPRequest http://www.webwareforpython.org/WebKit/Docs/Source/Docs/WebKit.HTTPRequest.html I try with webkit.NetworkRequest() but I don't know how to it. Some help? Thanks

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