Develop Reference

Delete DNS 'A' Records from Domain Controller using Python

I have a DC with "example.com" and I have many DNS records/FQDNs with 'A' record(Windows Servers).
Ex: Server1.example.com A 172.3.2.1
I'm using Python and trying to delete the record (server1).
Unfortunately, its giving me response as None.
I am using dnspython library.
def DeleteDNSRecords(serverList):
try:
for server in serverList:
updater = update.Update(server,'A')
response = updater.delete(server,'A')
print(str(response))
except Exception as e:
print (e)

Python: Access ftp like browsers do, with proxy

I want to access a ftp server, anonymous, just for download. My company have a proxy, and ftp ports (21) are blocked. I can't access the ftp server directly.
What I whant to do is to write some code that behaves exactly the same way browsers do. The idea is that, if I can download the files with my browser, there is way to do it with code.
My code works when I try to access a web site outside the company, but is still not working for ftp servers.
proxy = urllib2.ProxyHandler({'https': 'proxy.mycompanhy.com:8080',
'http': 'proxy.mycompanhy.com:80',
'ftp': 'proxy.mycompanhy.com:21' })
auth = urllib2.HTTPBasicAuthHandler()
opener = urllib2.build_opener(proxy, auth, urllib2.HTTPHandler)
urllib2.install_opener(opener)
urlAddress = 'https://python.org'
# urlAddress = 'ftp://ftp1.cptec.inpe.br'
conn = urllib2.urlopen(urlAddress)
return_str = conn.read()
print return_str
When I try to access python.org, it works fine. If I remove the install_opener part, it does not work anymore, proving that the proxy is required.
When I use the ftp url, it blocks (or timeout if I choose to use these parameters).
I understand that ftp and http are two very different protocols.
What I don't understand is the mechanism that browsers use to access these ftp servers.
I mean, I don't know if there is a layer on server side that interfaces between http and ftp, retriveing a html; or if browser, in some other maner, access the ftp and builds the page.
There also might be a confusion with the ftp domain (or the url) and the connection mode. It seems to me that when urllib2 reads the ftp://... it automatically uses the port 21.

I found a solution using wget. This package handles with proxies, but documentation was very ubscure. You need to setup an environment variable with proxy name.
import wget
import os
import errno
# setup proxy
os.environ["ftp_proxy"] = "proxy.mycompanhy.com"
os.environ["http_proxy"] = "proxy.mycompanhy.com"
os.environ["https_proxy"] = "proxy.mycompanhy.com"
src = "http://domain.gov/data/fileToDownload.txt"
out = "C:\\outFolder\\outFileName.txt" # out is optional
# create output folder if it doesn't exists
outFolder, _ = os.path.split( out )
try:
os.makedirs(outFolder)
except OSError as exc: # Python >2.5
if exc.errno == errno.EEXIST and os.path.isdir(outFolder):
pass
else: raise
# download
filename = wget.download(src, out)

Python - try statement breaking urllib2.urlopen

I'm writing a program in Python that has to make a http request while being forced onto a direct connection in order to avoid a proxy. Here is the code I use which successfully manages this:
print "INFO: Testing API..."
proxy = urllib2.ProxyHandler({})
opener = urllib2.build_opener(proxy)
urllib2.install_opener(opener)
req = urllib2.urlopen('http://maps.googleapis.com/maps/api/geocode/json?address=blahblah&sensor=true')
returneddata = json.loads(req.read())
I then want to add a try statement around 'req', in order to handle a situation where the user is not connected to the internet, which I have tried like so:
try:
req = urllib2.urlopen('http://maps.googleapis.com/maps/api/geocode/json?address=blahblah&sensor=true')
except urllib2.URLError:
print "Unable to connect etc etc"
The trouble is that by doing that, it always throws the exception, even though the address is perfectly accessible & the code works without it.
Any ideas? Cheers.

Feedparser (and urllib2) issue: Connection timed out

Starting with urllib2 and feedparser libraries in Python I'm getting the following error most of the time whenever try to connect and fetch content from particular URL:
urllib2.URLError: <urlopen error [Errno 110] Connection timed out>
The minimal reproducible examples (basic, using feedparser.parser directly and advanced, where I use urllib2 library first to fetch XML content) are pasted below.
# test-1
import feedparser
f = feedparser.parse('http://www.zurnal24.si/index.php?ctl=show_rss')
title = f['channel']['title']
print title
# test-2
import urllib2
import feedparser
url = 'http://www.zurnal24.si/index.php?ctl=show_rss'
opener = urllib2.build_opener()
opener.addheaders = [('User-Agent', 'Mozilla/5.0')]
request = opener.open(url)
response = request.read()
feed = feedparser.parse(response)
title = feed['channel']['title']
print title
When I try with different URL addresses (e.g., http://www.delo.si/rss/), everything works fine. Please note that all URL's lead to non-english (i.e., Slovenian) RSS feeds.
I run my experiments both from local and remote machine (via ssh). The error reported occurs more frequently on remote machine, although it's unpredictable even on local host.
Any suggestions would be greatly appreciated.

How often does the timeout occur? If it's not frequent, you could wait after each timeout and then retry the request:
import urllib2
import feedparser
import time
import sys
url = 'http://www.zurnal24.si/index.php?ctl=show_rss'
opener = urllib2.build_opener()
opener.addheaders = [('User-Agent', 'Mozilla/5.0')]
# Try to connect a few times, waiting longer after each consecutive failure
MAX_ATTEMPTS = 8
for attempt in range(MAX_ATTEMPTS):
try:
request = opener.open(url)
break
except urllib2.URLError, e:
sleep_secs = attempt ** 2
print >> sys.stderr, 'ERROR: %s.\nRetrying in %s seconds...' % (e, sleep_secs)
time.sleep(sleep_secs)
response = request.read()
feed = feedparser.parse(response)
title = feed['channel']['title']
print title

As the error denotes, it is a connection problem. This may be a problem with your internet connection or with their servers/connection/bandwidth..
A simple workaround is to do your feedparsing in a while loop, of course keeping a counter of MAX retries..

Python problems with FancyURLopener, 401, and "Connection: close"

I'm new to Python, so forgive me if I am missing something obvious.
I am using urllib.FancyURLopener to retrieve a web document. It works fine when authentication is disabled on the web server, but fails when authentication is enabled.
My guess is that I need to subclass urllib.FancyURLopener to override the get_user_passwd() and/or prompt_user_passwd() methods. So I did:
class my_opener (urllib.FancyURLopener):
# Redefine
def get_user_passwd(self, host, realm, clear_cache=0):
print "get_user_passwd() called; host %s, realm %s" % (host, realm)
return ('name', 'password')
Then I attempt to open the page:
try:
opener = my_opener()
f = opener.open ('http://1.2.3.4/whatever.html')
content = f.read()
print "Got it: ", content
except IOError:
print "Failed!"
I expect FancyURLopener to handle the 401, call my get_user_passwd(), and retry the request.
It does not; I get the IOError exception when I call "f = opener.open()".
Wireshark tells me that the request is sent, and that the server is sending a "401 Unauthorized" response with two headers of interest:
WWW-Authenticate: BASIC
Connection: close
The connection is then closed, I catch my exception, and it's all over.
It fails the same way even if I retry the "f = opener.open()" after IOError.
I have verified that my my_opener() class is working by overriding the http_error_401() method with a simple "print 'Got 401 error'". I have also tried to override the prompt_user_passwd() method, but that doesn't happen either.
I see no way to proactively specify the user name and password.
So how do I get urllib to retry the request?
Thanks.

I just tried your code on my webserver (nginx) and it works as expected:
Get from urllib client
HTTP/1.1 401 Unauthorized from server with Headers
Connection: close
WWW-Authenticate: Basic realm="Restricted"
client tries again with Authorization header
Authorization: Basic <Base64encoded credentials>
Server responds with 200 OK + Content
So I guess your code is right (I tried it with python 2.7.1) and maybe the webserver you are trying to access is not working as expected. Here is the code tested using the free http basic auth testsite browserspy.dk (seems they are using apache - the code works as expected):
import urllib
class my_opener (urllib.FancyURLopener):
# Redefine
def get_user_passwd(self, host, realm, clear_cache=0):
print "get_user_passwd() called; host %s, realm %s" % (host, realm)
return ('test', 'test')
try:
opener = my_opener()
f = opener.open ('http://browserspy.dk/password-ok.php')
content = f.read()
print "Got it: ", content
except IOError:
print "Failed!"