import urllib

import urllib2

from lxml import objectify
import sys


username = "USERNAME"
password = "PASSWORD"
ip_address = "192.168.5.146"


#yup, here is the password in plain text encoded in the url string, this needs to be cleaned up

url = 'http://' + ip_address + '/System.xml?version=1.0&action=login&userName=' + username + '&password=' + password

req = urllib2.Request(url)

response = urllib2.urlopen(req)

page = response.read()



#parse the resulting XML to see if the login worked. lxml might be overkill, but I like it

try:

    camRespond = objectify.fromstring(page)

except Error, e:

    print "Error",e



if camRespond.statusCode == 1:

    print "Login success"

    #the sessionID is returned in in the responding http header

    info = response.info()

    sessionID = info.get("sessionID")

    print "sessionID =", sessionID

else:

    print "login failed!"

    print camRespond
    sys.exit(1)



url = 'http://' + ip_address + '/StreamingSetting.xml?version=1.0&action=get&sessionID=' + sessionID

req = urllib2.Request(url)

response = urllib2.urlopen(req)

page = response.read()

print page





#logout and clean up

#url = 'http://' + ip_address + '/System.xml?version=1.0&action=logout&sessionID=' + sessionID

#req = urllib2.Request(url)

#response = urllib2.urlopen(req)

#page = response.read()

#print page