Hi i am able to parse a normal xml like xml = lxml.etree.parse(''http://abc.com/A.xml')
but now i have this path authenticated with a user name and password is it possible to input the username and password and parse the url, like in connecting a database where you can give the user name password in the connection string
Yes, it's possible. Before parsing the XML document with lxml, you need to get it by making an HTTP request that handles the HTTP Basic/Digest Authentication properly. For example, with urllib2.HTTPBasicAuthHandler like in this solution: Python urllib2 HTTPBasicAuthHandler
Guys i found a way to parse password protected XML this is what i did.
import urllib2
import base64
theurl = 'http://abc.com/A.xml'
username='AAA'
password='BBB'
req = urllib2.Request(theurl)
base64string = base64.encodestring(
'%s:%s' % (username, password))[:-1]
authheader = "Basic %s" % base64string
req.add_header("Authorization", authheader)
try:
handle = urllib2.urlopen(req)
except IOError, e:
print "It looks like the username or password is wrong."
xml = handle.read()
inputXml = etree.fromstring(xml)
Related
I am trying write a python script using requests package to use an online mongodb query service API hosted within the organization. The API expects the authorization header in the format 'websitename/username:Password' and using the basic authentication base64 encoding. I tried to create the GET request using the requests package which has the authorization header in the following format:
import requests
headers = {'Authorization': 'Basic %s' % 'Base64encoded
websitename/username:Password string here'}
content_res = requests.get(get_url, headers = headers).json()
But I am getting a parse error here for the string as I think the expected string for header in requests package is in form of 'username:password' here and not the desired format i.e. 'websitename/username:password'.
Is there a way in which I could use the base64 encoded sting in the format which the service is expecting i.e. 'websitename/username:password' in requests package?
Any help is highly appreciated.
Thanks.
It sounds to me like you are getting a HTTP response error because the authorization header value you are passing is not base64 encoded. To correct this you can simply encode the string using the base64 python module:
Python 2.7 https://docs.python.org/2/library/base64.html
Python 3.5 https://docs.python.org/3.5/library/base64.html
An example would be something like this:
import base64
import requests
website = 'example.com'
username = 'root'
password = '1234'
auth_str = '%s/%s:%s' % (website, username, password)
b64_auth_str = base64.b64encode(auth_str)
headers = {'Authorization': 'Basic %s' % b64_auth_str}
content_res = requests.get(get_url, headers=headers).json()
import base64
import requests
website = 'example.com'
username = 'root'
password = '1234'
auth_str = '%s/%s:%s' % (website, username, password)
b64_auth_str = base64.b64encode(auth_str.encode('ascii'))
headers = {'Authorization': 'Basic %s' % b64_auth_str}
content_res = requests.get(get_url, headers=headers).json()
I tried to call AppDynamics API using python requests but face an issue.
I wrote a sample code using the python client as follows...
from appd.request import AppDynamicsClient
c = AppDynamicsClient('URL','group','appd#123')
for app in c.get_applications():
print app.id, app.name
It works fine.
But if I do a simple call like the following
import requests
usr =<uid>
pwd =<pwd>
url ='http://10.201.51.40:8090/controller/rest/applications?output=JSON'
response = requests.get(url,auth=(usr,pwd))
print 'response',response
I get the following response:
response <Response [401]>
Am I doing anything wrong here ?
Couple of things:
I think the general URL format for app dynamics applications are (notice the '#'):
url ='http://10.201.51.40:8090/controller/#/rest/applications?output=JSON'
Also, I think the requests.get method needs an additional parameter for the 'account'. For instance, my auth format looks like:
auth = (_username + '#' + _account, _password)
I am able to get a right response code back with this config. Let me know if this works for you.
You could also use native python code for more control:
example:
import os
import sys
import urllib2
import base64
# if you have a proxy else comment out this line
proxy = urllib2.ProxyHandler({'https': 'proxy:port'})
opener = urllib2.build_opener(proxy)
urllib2.install_opener(opener)
username = "YOUR APPD REST API USER NAME"
password = "YOUR APPD REST API PASSWORD"
#Enter your request
request = urllib2.Request("https://yourappdendpoint/controller/rest/applications/141/events?time-range-type=BEFORE_NOW&duration-in-mins=5&event-types=ERROR,APPLICATION_ERROR,DIAGNOSTIC_SESSION&severities=ERROR")
base64string = base64.encodestring('%s:%s' % (username, password)).replace('\n', '')
request.add_header("Authorization", "Basic %s" % base64string)
response = urllib2.urlopen(request)
html = response.read()
This will get you the response and you can parse the XML as needed.
If you prefer it in JSON simply specify it in the request.
Hello I'm making a python script for geektool to monitor my incoming mobile messages.
I've succeed to make it in java but i've never used python.
api url: https://mobilevikings.com/api/2.0/basic/usage.json
password: xxxx
username: yyyy
i already have the following non-working code, copy pasted from the internet
import urllib2, base64
request = urllib2.Request(url)
base64string = base64.encodestring('%s:%s' % (username, password)).replace('\n', '')
request.add_header("Authorization", "Basic %s" % base64string)
result = urllib2.urlopen(request)
How do i receive the json file from the api?
I have VMware setup for testing. I create one user abc/abc123 to access the Org url "http://localhost/cloud/org/MyOrg". I want to access the RestAPI of the VCloud. I tried with RestClient plugin in firefox. Its working fine.
Now I tried with python code.
url = 'https://localhost/api/sessions/'
req = urllib2.Request(url)
base64string = base64.encodestring('%s:%s' % ('abc#MyOrg', 'abc123'))[:-1]
authheader = "Basic %s" % base64string
req.add_header("Authorization", authheader)
req.add_header("Accept", 'application/*+xml;version=1.5')
f = urllib2.urlopen(req)
data = f.read()
print(data)
This is the code i get from stackoverflow. But for my example its give "urllib2.HTTPError: HTTP Error 403: Forbidden" Error.
I also tried HTTP authentication for the same.
After doing some googling I found the solution from the post https://stackoverflow.com/a/6348729/243031. I change the code for my usability. I am posting the answer because if some one has same error then he will get the answer directly.
My change code is:
import urllib2
import base64
# make a string with the request type in it:
method = "POST"
# create a handler. you can specify different handlers here (file uploads etc)
# but we go for the default
handler = urllib2.HTTPSHandler()
# create an openerdirector instance
opener = urllib2.build_opener(handler)
# build a request
url = 'https://localhost/api/sessions'
request = urllib2.Request(url)
# add any other information you want
base64string = base64.encodestring('%s:%s' % ('abc#MyOrg', 'abc123'))[:-1]
authheader = "Basic %s" % base64string
request.add_header("Authorization", authheader)
request.add_header("Accept",'application/*+xml;version=1.5')
# overload the get method function with a small anonymous function...
request.get_method = lambda: method
# try it; don't forget to catch the result
try:
connection = opener.open(request)
except urllib2.HTTPError,e:
connection = e
# check. Substitute with appropriate HTTP code.
if connection.code == 200:
data = connection.read()
print "Data :", data
else:
print "ERRROR", connection.code
Hope this will help some one who want to send POST request without the data.
How do you open https url in Python?
import urllib2
url = "https://user:password#domain.com/path/
f = urllib2.urlopen(url)
print f.read()
gives:
httplib.InvalidURL: nonnumeric port: 'password#domain.com'
This has never failed me
import urllib2, base64
username = 'foo'
password = 'bar'
auth_encoded = base64.encodestring('%s:%s' % (username, password))[:-1]
req = urllib2.Request('https://somewebsite.com')
req.add_header('Authorization', 'Basic %s' % auth_encoded)
try:
response = urllib2.urlopen(req)
except urllib2.HTTPError, http_e:
# etc...
pass
Please read about the urllib2 password manager and the basic authentication handler as well as the digest authentication handler.
http://docs.python.org/library/urllib2.html#abstractbasicauthhandler-objects
http://docs.python.org/library/urllib2.html#httpdigestauthhandler-objects
Your urllib2 script must actually provide enough information to do HTTP authentication. Usernames, Passwords, Domains, etc.
If you want to pass username and password information to urllib2 you'll need to use an HTTPBasicAuthHandler.
Here's a tutorial showing you how to do it.
You cannot pass credentials to urllib2.open like that. In your case, user is interpreted as the domain name, while password#domain.com is interpreted as the port number.