Open a XML to send to SOAP in Python

Question

I am implementing a client for interaction with a SOAP WebService. This WebService has two parameters: a string and an xml.
I am a beginner in development with python but I would like to know how I can get an xml file from my machine to send it to my client.
I Tried :

test = open('LCL.XML')  
test = test.read()
print(test)

But it did not work.

return codecs.charmap_decode(input,self.errors,decoding_table)[0] UnicodeDecodeError: 'charmap' codec can't decode byte 0x81 in position
 492: character maps to <undefined>

I'm a Perl guy, but a SOAP person, so I'd recommend using a library, for example: http://docs.python-zeep.org/en/master/ this will make things simpler — Hugh Barnard, Apr 03 '17 at 12:06
I'm using zeep for a SOAP request. But my question is about open a xml file. Thanks — NTI, Apr 03 '17 at 12:13
Use a `xml module`, for instance: https://docs.python.org/2/library/xml.etree.elementtree.html#parsing-xml — stovfl, Apr 03 '17 at 14:10
http://stackoverflow.com/questions/14242486/python-why-am-i-getting-a-unicodedecodeerror may help. — Hugh Barnard, Apr 04 '17 at 16:07

score 0 · Answer 1 · answered Sep 06 '17 at 14:37

Here is how to read and xml file and get it ready for sending vis SOAP:

import xml.etree.ElementTree filename = 'whatever.xml' result = xml.etree.ElementTree.parse(filename).getroot() xml_str = xml.etree.ElementTree.tostring(result, encoding='utf-8')

Then send xml_str in zeep.Client.service.<your soap call>(..., xml_str, ...)

Open a XML to send to SOAP in Python

1 Answers1