python xmlschema 标识所有嵌套属性

Question

I am trying to traverse through an xml schema. The goal is to print out all the complextypes with their nested components. This xsd has nested complextypes.

I am using 'xmlschema' to do this. In the below code I could only pick out the complextypes in the begining but in order to drill down I need to take actions based on the element\validator type. I can't seem to specify a statement which says if it is a complextype do this or if it is simpletype then do something else:

import xmlschema

from pprint import pprint
schema = xmlschema.XMLSchema('Schema.xsd')
for t in schema.complex_types:
    print(t.name)
    for c in t.iter_components():
        print('\t', c.name)
        print('\t', type(c))

Answer 1

To traverse all the complex types defined (globals and locals) you can iterate components from the schema instance and use the option that filters the XSD component types returned:

import xmlschema
from xmlschema.validators import XsdComplexType

schema = xmlschema.XMLSchema('Schema.xsd')
for xsd_type in t.iter_components(xsd_classes=XsdComplexType):
    print(xsd_type)