API Calls with Python 3

Read 101 times
I'm back again after getting the XML API working with Python2. I need Python3's urllib3 to read gzipped RSS feeds and then collect podcast episodes and upload them to my Centova server.

So far so good, I can collect the episodes and get them staged, tagged and ready. What's failing is any attempt to talk to the Centova server via the api.

Here's what I have:

import urllib3

def addToPlaylist(playlist,fileToAdd):
    # add the file to the scheduled show
    additem =  '<?xml version="1.0" encoding="UTF-8"?>\
    <request class="server" method="playlist">\
    </centovacast>' % (playlist,fileToAdd)

    URL = 'http://mystation.com:2199/api.php?'
    pageEngineForAdding = urllib3.PoolManager()

    response = pageEngineForAdding.request('GET',URL,)

    result = str(response.data,encoding='utf-8')

What I'm sending as a URL is:
http://mystation.com:2199/api.php?<?xml version="1.0" encoding="UTF-8"?>    <centovacast>    <request class="server" method="playlist">    <action>add</action>    <playlistname>The Two Gay Geeks 3 pm Tue</playlistname>    <trackpath>thetwogaygeeks_prerecorded/TG-Geeks-Webcast-Episode-201.mp3</trackpath>    <password>my_password</password>    <username>my_username</username>    </request>    </centovacast>

And what I get back as an error code when this runs is:

<?xml version="1.0" encoding="UTF-8"?><centovacast version="3.2.10" host="mystation.com:2199"><response type="error"><message>XML root tag not found.</message></response></centovacast>

Anybody have any ideas on what might be happening on this, and what I might do to fix this?