Right way for handling xsd:any response with zeep
I'm using zeep for retrieving data from SOAP service. In my case response is an xsd:any and zeep have no idea what is it, so it just returns lxml.element (explanation on github)
What is the best way to handle this case?
This is how I'm trying to do it now:
from io import StringIO
import xml.etree.ElementTree as ET
# define a couple of helpers
def strip_namespaces(xml):
"""
parses raw xml and removes namespaces from tags.
"""
iterable = ET.iterparse(StringIO(xml))
for _, el in iterable:
if '}' in el.tag:
el.tag = el.tag.split('}', 1)[1] # strip all namespaces
root = iterable.root
return root
def element_to_dict(element):
"""converts lxml object into python dict"""
return (element.tag,
dict(map(element_to_dict, element)) or element.text)
with client.settings(raw_response=True): # ask zeep to don't parse response
result = client.service.Method(argument)
el = strip_namespaces(result.text)
_, response_dict = element_to_dict(el)
print(response_dict)
It works (I guess), but looks a bit verbose and fragile.
Is there a more "pythonic" way?
python zeep
asked Nov 15 '18 at 14:33
NobbyNobbsNobbyNobbs
3631314
add a comment |
I'm using zeep for retrieving data from SOAP service. In my case response is an xsd:any and zeep have no idea what is it, so it just returns lxml.element (explanation on github)
What is the best way to handle this case?
This is how I'm trying to do it now:
from io import StringIO
import xml.etree.ElementTree as ET
# define a couple of helpers
def strip_namespaces(xml):
"""
parses raw xml and removes namespaces from tags.
"""
iterable = ET.iterparse(StringIO(xml))
for _, el in iterable:
if '}' in el.tag:
el.tag = el.tag.split('}', 1)[1] # strip all namespaces
root = iterable.root
return root
def element_to_dict(element):
"""converts lxml object into python dict"""
return (element.tag,
dict(map(element_to_dict, element)) or element.text)
with client.settings(raw_response=True): # ask zeep to don't parse response
result = client.service.Method(argument)
el = strip_namespaces(result.text)
_, response_dict = element_to_dict(el)
print(response_dict)
It works (I guess), but looks a bit verbose and fragile.
Is there a more "pythonic" way?
python zeep
asked Nov 15 '18 at 14:33
NobbyNobbsNobbyNobbs
3631314
add a comment |
I'm using zeep for retrieving data from SOAP service. In my case response is an xsd:any and zeep have no idea what is it, so it just returns lxml.element (explanation on github)
What is the best way to handle this case?
This is how I'm trying to do it now:
from io import StringIO
import xml.etree.ElementTree as ET
# define a couple of helpers
def strip_namespaces(xml):
"""
parses raw xml and removes namespaces from tags.
"""
iterable = ET.iterparse(StringIO(xml))
for _, el in iterable:
if '}' in el.tag:
el.tag = el.tag.split('}', 1)[1] # strip all namespaces
root = iterable.root
return root
def element_to_dict(element):
"""converts lxml object into python dict"""
return (element.tag,
dict(map(element_to_dict, element)) or element.text)
with client.settings(raw_response=True): # ask zeep to don't parse response
result = client.service.Method(argument)
el = strip_namespaces(result.text)
_, response_dict = element_to_dict(el)
print(response_dict)
It works (I guess), but looks a bit verbose and fragile.
Is there a more "pythonic" way?
python zeep
asked Nov 15 '18 at 14:33
NobbyNobbsNobbyNobbs
3631314
I'm using zeep for retrieving data from SOAP service. In my case response is an xsd:any and zeep have no idea what is it, so it just returns lxml.element (explanation on github)
What is the best way to handle this case?
This is how I'm trying to do it now:
from io import StringIO
import xml.etree.ElementTree as ET
# define a couple of helpers
def strip_namespaces(xml):
"""
parses raw xml and removes namespaces from tags.
"""
iterable = ET.iterparse(StringIO(xml))
for _, el in iterable:
if '}' in el.tag:
el.tag = el.tag.split('}', 1)[1] # strip all namespaces
root = iterable.root
return root
def element_to_dict(element):
"""converts lxml object into python dict"""
return (element.tag,
dict(map(element_to_dict, element)) or element.text)
with client.settings(raw_response=True): # ask zeep to don't parse response
result = client.service.Method(argument)
el = strip_namespaces(result.text)
_, response_dict = element_to_dict(el)
print(response_dict)
It works (I guess), but looks a bit verbose and fragile.
Is there a more "pythonic" way?
python zeep
python zeep
asked Nov 15 '18 at 14:33
NobbyNobbsNobbyNobbs
3631314
asked Nov 15 '18 at 14:33
NobbyNobbsNobbyNobbs
3631314
edited Nov 15 '18 at 14:38
NobbyNobbs
asked Nov 15 '18 at 14:33
NobbyNobbsNobbyNobbs
3631314
asked Nov 15 '18 at 14:33
NobbyNobbsNobbyNobbs
3631314
asked Nov 15 '18 at 14:33
NobbyNobbsNobbyNobbs
3631314
3631314
add a comment |
add a comment |
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