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Partial response of documentconversionV1()

I am trying to use DocumentConversionV1 function of watson_developer_cloud API on python , However the response in my case comes only as "<"Response 200">".
import sys
import os as o
import json
import codecs
from watson_developer_cloud import DocumentConversionV1
document_conversion = DocumentConversionV1(
username="873512ac-dcf7-4365-a01d-7dec438d5720",
password="bvhXbdaHtYgw",
version='2016-02-10',
url= "https://gateway.watsonplatform.net/document-conversion/api",
)
config = {
'conversion_target': 'NORMALIZED_TEXT',
}
i = "v.docx"
with open((i),'rb') as doc:
res = document_conversion.convert_document(document = doc , config = config)
print(res)

First and foremost, delete your service credentials and recreate them through Bluemix. (Posting them on a public forum is usually a bad idea.) ;o)
Now, to actually answer the question... You want to get the content of the response. Right now, you're printing the response itself. Try
print(res.content)
See https://github.com/watson-developer-cloud/python-sdk/blob/master/examples/document_conversion_v1.py#L16

Weird json value urllib python

I'm trying to manipulate a dynamic JSON from this site:
http://esaj.tjsc.jus.br/cposgtj/imagemCaptcha.do
It has 3 elements, imagem, a base64, labelValorCaptcha, just a message, and uuidCaptcha, a value to pass by parameter to play a sound in this link bellow:
http://esaj.tjsc.jus.br/cposgtj/somCaptcha.do?timestamp=1455996420264&uuidCaptcha=sajcaptcha_e7b072e1fce5493cbdc46c9e4738ab8a
When I enter in the first site through a browser and put in the second link the uuidCaptha after the equal ("..uuidCaptcha="), the sound plays normally. I wrote a simple code to catch this elements.
import urllib, json
url = "http://esaj.tjsc.jus.br/cposgtj/imagemCaptcha.do"
response = urllib.urlopen(url)
data = json.loads(response.read())
urlSound = "http://esaj.tjsc.jus.br/cposgtj/somCaptcha.do?timestamp=1455996420264&uuidCaptcha="
print urlSound + data['uuidCaptcha']
But I dont know what's happening, the caught value of the uuidCaptcha doesn't work. Open a error web page.
Someone knows?
Thanks!

It works for me.
$ cat a.py
#!/usr/bin/env python
# encoding: utf-8
import urllib, json
url = "http://esaj.tjsc.jus.br/cposgtj/imagemCaptcha.do"
response = urllib.urlopen(url)
data = json.loads(response.read())
urlSound = "http://esaj.tjsc.jus.br/cposgtj/somCaptcha.do?timestamp=1455996420264&uuidCaptcha="
print urlSound + data['uuidCaptcha']
$ python a.py
http://esaj.tjsc.jus.br/cposgtj/somCaptcha.do?timestamp=1455996420264&uuidCaptcha=sajcaptcha_efc8d4bc3bdb428eab8370c4e04ab42c

As I said #Charlie Harding, the best way is download the page and get the JSON values, because this JSON is dynamic and need an opened web link to exist.
More info here.

Strange Output from Python urllib2

I would like to read to source code of a webpage using urllib2; however, I'm seeing a strange output that I've not seen before. Here's the code (Python 2.7, Linux):
import urllib2
open_url = urllib2.urlopen("http://www.elegantthemes.com/gallery/")
site_html = open_url.read()
site_html[50:]
Which gives the output:
'\x1f\x8b\x08\x00\x00\x00\x00\x00\x00\x03\xe5\\ms\xdb\xb6\xb2\xfel\xcf\xe4?\xc0<S[\x9a\x8a\xa4^\xe28u,\xa5\x8e\x93\xf4\xa4\x93&\x99:9\xbdw\x9a\x8e\x07"'
Does anyone know why it's showing this as the output and not the correct HTML?

The http response being sent by the site is actually gzipped content and hence the strange output. urllib does not automatically decode the gzip cntent. There are two ways to solve this -
1) Decode zipped content before printing -
import urllib2
import io
import gzip
open_url = urllib2.urlopen("http://www.elegantthemes.com/gallery/")
site_html = open_url.read()
bi = io.BytesIO(site_html)
gf = gzip.GzipFile(fileobj=bi, mode="rb")
s = gf.read()
print s[50:]
2) Use Requests library -
import requests
r = requests.get('http://www.elegantthemes.com/gallery/')
print r.content

How do you IXR_Base64 in python?

What I'm trying to do is upload a picture to wordpress using wp.uploadFile xmlrpc method.
To do this, in PHP there is an example here: https://stackoverflow.com/a/8910496/1212382
I'm trying to do the same thing in python but I don't know how.
Anyone any ideas?

ok, the answer lies in the xmlrpclib class.
To send base64 bits to wordpress from python you need to use the xmlrpclib class like so:
base64bits = xmlrpclib.Binary(file_content)
then you just add the base64bits variable to the 'bits' parameter in your wp.uploadFile xmlrpc request.
to be a little more exact, here's the complete code in python of how this should be done:
import xmlrpclib
import urllib2
from datetime import date
import time
def get_url_content(url):
try:
content = urllib2.urlopen(url)
return content.read()
except:
print 'error! NOOOOOO!!!'
file_url = 'http://the path to your picture'
extension = file_url.split(".")
leng = extension.__len__()
extension = extension[leng-1]
if (extension=='jpg'):
xfileType = 'image/jpeg'
elif(extension=='png'):
xfileType='image/png'
elif(extension=='bmp'):
xfileType = 'image/bmp'
file = get_url_content(file_url)
file = xmlrpclib.Binary(file)
server = xmlrpclib.Server('http://website.com/xmlrpc.php')
filename = str(date.today())+str(time.strftime('%H:%M:%S'))
mediarray = {'name':filename+'.'+extension,
'type':xfileType,
'bits':file,
'overwrite':'false'}
xarr = ['1', 'USERHERE', 'PASSWORDHERE', mediarray]
result = server.wp.uploadFile(xarr)
print result

Python code like curl

in curl i do this:
curl -u email:password http://api.foursquare.com/v1/venue.json?vid=2393749
How i can do this same thing in python?

Here's the equivalent in pycurl:
import pycurl
from StringIO import StringIO
response_buffer = StringIO()
curl = pycurl.Curl()
curl.setopt(curl.URL, "http://api.foursquare.com/v1/venue.json?vid=2393749")
curl.setopt(curl.USERPWD, '%s:%s' % ('youruser', 'yourpassword'))
curl.setopt(curl.WRITEFUNCTION, response_buffer.write)
curl.perform()
curl.close()
response_value = response_buffer.getvalue()

"The problem could be that the Python libraries, per HTTP-Standard, first send an unauthenticated request, and then only if it's answered with a 401 retry, are the correct credentials sent. If the Foursquare servers don't do "totally standard authentication" then the libraries won't work.
Try using headers to do authentication:"
taked from Python urllib2 Basic Auth Problem
import urllib2
import base64
req = urllib2.Request('http://api.foursquare.com/v1/venue.json?vid=%s' % self.venue_id)
req.add_header('Authorization: Basic ',base64.b64encode('email:password'))
res = urllib2.urlopen(req)

I'm more comfortable running the command line curl through subprocess. This avoids all of the potential version matching headaches of python, pycurl, and libcurl. The observation that pycurl hasn't been touched in 2 years, and is only listed as suppported through Python 2.5, made me wary.
-- John
import subprocess
def curl(*args):
curl_path = '/usr/bin/curl'
curl_list = [curl_path]
for arg in args:
curl_list.append(arg)
curl_result = subprocess.Popen(
curl_list,
stderr=subprocess.PIPE,
stdout=subprocess.PIPE).communicate()[0]
return curl_result
answer = curl('-u', 'email:password', 'http://api.foursquare.com/v1/venue.json?vid=2393749')

if use human_curl you can write some code
import human_curl as hurl
r = hurl.get('http://api.foursquare.com/v1/venue.json?vid=2393749', auth=('email','password'))
Json data in r.content

Use pycurl
http://pycurl.sourceforge.net/
There is a discussion on SO for tutorials
What good tutorials exist for learning pycURL?
A typical example:
import sys
import pycurl
class ContentCallback:
def __init__(self):
self.contents = ''
def content_callback(self, buf):
self.contents = self.contents + buf
t = ContentCallback()
curlObj = pycurl.Curl()
curlObj.setopt(curlObj.URL, 'http://www.google.com')
curlObj.setopt(curlObj.WRITEFUNCTION, t.content_callback)
curlObj.perform()
curlObj.close()
print t.contents