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://localhost: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.
Okay, figured it out.
The urllib2.Request line needs to have a slash in front of the path so:-
request = urllib2.Request('http://localhost:8080/manager/deploy?path=/war_file', data=war_file_contents)
All then works fine.
Related
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.
I have a Python script used to connect to Parse.com (remote server) and upload a file. The script runs off a server that sits behind a corporate firewall.
import env
import json
import requests
from requests.auth import HTTPProxyAuth
def uploadFile(fileFullPath):
print "Attempting to upload file: " + fileFullPath
proxies = {
"http": "http://10.128.198.14",
"https": "http://10.128.198.14"
}
auth = HTTPProxyAuth('MyDomain\MyUsername', 'MyPassord')
headers = {
"X-Parse-Application-Id": env.X_Parse_APP_ID,
"X-Parse-REST-API-Key": env.X_Parse_REST_API_Key,
"Content-Type": "application/pdf"
}
f = open(fileFullPath, 'r')
files = {'file': f}
r = requests.post(env.PARSE_HOSTNAME + env.PARSE_FILES_ENDPOINT + "/" + env.PARSE_FILE_NAME, files=files, headers=headers, timeout=10, verify=False, proxies=proxies)
print r.text
When I used this module from the command prompt, I got the following message:
ConnectionError thrown. Details: Cannot connect to proxy. Socket error: Tunnel connection failed: 407 Proxy Authentication Required.
I am pretty sure the username and password are both correct.
Any solution? Thanks!
The reason for the 407 error is that the proxy itself needs to be authenticated. So for your proxies dict, do the following:
proxies = {
"http": "http://user:pass#10.128.198.14",
"https": "http://user:pass#10.128.198.14"
}
Fill in the user and pass variables in the proxies urls. Here is a link to the relevant requests documentation on how to build proxy objects and have them authenticated.
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 want to access Twitter 1.1 search endpoint using application-only authentication. To do the same, I'm trying to implement the steps given on Twitter API's documentation here - https://dev.twitter.com/docs/auth/application-only-auth (scroll to "Issuing application-only requests")
I am not able to obtain the "bearer token" in Step 2. When I run the following code, I receive "Response: 302 Found" which is a redirection to Location: https://api.twitter.com/oauth2/token
Ideally it should be "200 OK"
import urllib
import base64
import httplib
CONSUMER_KEY = 'my_key'
CONSUMER_SECRET = 'my_secret'
encoded_CONSUMER_KEY = urllib.quote(CONSUMER_KEY)
encoded_CONSUMER_SECRET = urllib.quote(CONSUMER_SECRET)
concat_consumer_url = encoded_CONSUMER_KEY + ":" + encoded_CONSUMER_SECRET
host = 'api.twitter.com'
url = '/oauth2/token'
params = urllib.urlencode({'grant_type' : 'client_credentials'})
req = httplib.HTTP(host)
req.putrequest("POST", url)
req.putheader("Host", host)
req.putheader("User-Agent", "My Twitter 1.1")
req.putheader("Authorization", "Basic %s" % base64.b64encode(concat_consumer_url))
req.putheader("Content-Type" ,"application/x-www-form-urlencoded;charset=UTF-8")
req.putheader("Content-Length", "29")
req.putheader("Accept-Encoding", "gzip")
req.endheaders()
req.send(params)
# get the response
statuscode, statusmessage, header = req.getreply()
print "Response: ", statuscode, statusmessage
print "Headers: ", header
I do not want to use any Twitter API wrappers to access this.
The problem was that the URL had to be called with an HTTPS connection. Please check the modified code which works.
import urllib
import base64
import httplib
CONSUMER_KEY = 'my_key'
CONSUMER_SECRET = 'my_secret'
encoded_CONSUMER_KEY = urllib.quote(CONSUMER_KEY)
encoded_CONSUMER_SECRET = urllib.quote(CONSUMER_SECRET)
concat_consumer_url = encoded_CONSUMER_KEY + ":" + encoded_CONSUMER_SECRET
host = 'api.twitter.com'
url = '/oauth2/token/'
params = urllib.urlencode({'grant_type' : 'client_credentials'})
req = httplib.HTTPSConnection(host)
req.putrequest("POST", url)
req.putheader("Host", host)
req.putheader("User-Agent", "My Twitter 1.1")
req.putheader("Authorization", "Basic %s" % base64.b64encode(concat_consumer_url))
req.putheader("Content-Type" ,"application/x-www-form-urlencoded;charset=UTF-8")
req.putheader("Content-Length", "29")
req.putheader("Accept-Encoding", "gzip")
req.endheaders()
req.send(params)
resp = req.getresponse()
print resp.status, resp.reason
Although this is a bit late you might find this github page of some help. I've started creating a library for twitter application only authentication methods.
http://jonhurlock.github.io/Twitter-Application-Only-Authentication-OAuth-Python/
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.