Basically I have this xml element (xml.etree.ElementTree) and I want to POST it to a url. Currently I'm doing something like
xml_string = xml.etree.ElementTree.tostring(my_element)
data = urllib.urlencode({'xml': xml_string})
response = urllib2.urlopen(url, data)
I'm pretty sure that works and all, but was wondering if there is some better practice or way to do it without converting it to a string first.
Thanks!
If this is your own API, I would consider POSTing as application/xml. The default is application/x-www-form-urlencoded, which is meant for HTML form data, not a single XML document.
req = urllib2.Request(url=url,
data=xml_string,
headers={'Content-Type': 'application/xml'})
urllib2.urlopen(req)
urllib2.urlopen(req) -- urlopen can take Request objects as well as plain URL strings.Here is a full example (snippet) for sending post data (xml) to an URL:
def execQualysAction(username,password,url,request_data):
import urllib,urrlib2
xml_output = None
try:
base64string = base64.encodestring('%s:%s' % (username, password)).replace('\n', '')
headers = {'X-Requested-With' : 'urllib2','Content-Type': 'application/xml','Authorization': 'Basic %s' % base64string}
req = urllib2.Request(url=url,data=request_data,headers=headers)
response = urllib2.urlopen(req,timeout=int(TIMEOUT))
xml_output = response.read()
if args.verbose>1:
print "Result of executing action request",request_data,"is:",xml_output
except:
xml_output = '<RESULT></RESULT>'
traceback.print_exc(file=sys.stdout)
print '-'*60
finally:
return xml_output
No, I think that's probably the best way to do it - it's short and simple, what more could you ask for? Obviously the XML has to be converted to a string at some point, and unless you're using an XML library with builtin support for POSTing to a URL (which xml.etree is not), you'll have to do it yourself.
