CODEWITHSUNDEEP
pythonxmlwsdlzeep
Marc
Marc

Reputation: 3596

How to extract XML from zeep response

Suppose I use python zeep to execute a query against a server like this:

from zeep import Client

wsdl_url = "http://webservices.oorsprong.org/websamples.countryinfo/CountryInfoService.wso?WSDL"

client = Client(wsdl_url)
result = client.service.ListOfContinentsByCode()

I would like to see the raw XML that comes back from this method, ie extract it somehow from result. Is this possible? How?

Upvotes: 6

Views: 13254

Answers (3)

kjpc-tech
kjpc-tech

Reputation: 570

If you are trying to debug the XML in a zeep request or response you can use standard Python logging. From the documentation:

To see the SOAP XML messages which are sent to the remote server and the response received you can set the Python logger level to DEBUG for the zeep.transports module.

logging.config.dictConfig({
    ...
    'loggers': {
        'zeep.transports': {
            'level': 'DEBUG',
            'propagate': True,
            'handlers': ['console'],
        },
    },
})

Upvotes: 3

Trapli
Trapli

Reputation: 1597

I think what you are looking for is the history plugin.

from zeep.plugins import HistoryPlugin
from zeep import Client
from lxml import etree

wsdl_url = "http://webservices.oorsprong.org/websamples.countryinfo/CountryInfoService.wso?WSDL"

history = HistoryPlugin()
client = Client(wsdl_url, plugins=[history])
client.service.CurrencyName('GBP')
your_pretty_xml = etree.tostring(history.last_received["envelope"], encoding="unicode", pretty_print=True)

Upvotes: 8

balderman
balderman

Reputation: 23825

Here (requests takes care of the gzip)

import requests
r = equests.get('http://webservices.oorsprong.org/websamples.countryinfo/CountryInfoService.wso?WSDL')
if r.status_code == 200:
    print(r.content)

Upvotes: -3

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