Parse an XSD file using python

问题

I am trying to generate an XML file from a given XML schema. I have been able to do it with pyxb library in python. But the problem is as the XSD gets huge it is impossible to manually encode each and evey tag. Is there any python library which creates a data structure from a given XSD file which can be iterated through

回答1:

You can generate XML file from XSD file:

import requests

with open('file.xsd', 'r') as f:
    data = f.read()
r = requests.post('https://www.liquid-technologies.com/api/Converter', json={"Filename": "schema.xsd", "Type": "xsd", "Data": data, "TargetType": "xml", "Arguments": {"elementName": "Root", "elementNamespace": "", "addAnyAttributes": False, "addAnyElements": False, "forceOptionItemsToDepthOf": "4", "forceXsiNamespaceDelaration": False, "maxDepthToCreateOptionalItems": "7", "indent": "2", "indentChar": " ", "indentAttributes": False, "seed": "9722"}})
with open('file.xml', 'w') as f:
    f.write(r.json()['Files'][0]['Data'])

回答2:

This library seems to do what you want: https://pypi.org/project/xmlschema/

After skimming the documentation I have found this code example: https://xmlschema.readthedocs.io/en/latest/usage.html#xsd-declarations

>>> import xmlschema
>>> from pprint import pprint
>>> schema = xmlschema.XMLSchema('xmlschema/tests/test_cases/examples/vehicles/vehicles.xsd')
>>> schema.types
NamespaceView({'vehicleType': XsdComplexType(name='vehicleType')})
>>> pprint(dict(schema.elements))
{'bikes': XsdElement(name='vh:bikes', occurs=[1, 1]),
 'cars': XsdElement(name='vh:cars', occurs=[1, 1]),
 'vehicles': XsdElement(name='vh:vehicles', occurs=[1, 1])}
>>> schema.attributes
NamespaceView({'step': XsdAttribute(name='vh:step')})

So it looks like it can be used to create a python data-structure you can iterate through from an XSD file.

Also this question might be relevant: How to convert XSD to Python Class

来源：https://stackoverflow.com/questions/54687837/parse-an-xsd-file-using-python