How to POST an xml element in python

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说谎
说谎 2021-02-01 08:32

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         


        
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  • 2021-02-01 09:10

    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
    
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  • 2021-02-01 09:15

    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.

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  • 2021-02-01 09:18

    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)
    
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