I need to parse a file .xsd in Python as i would parse an XML.
I am using libxml2.
I have to parse an xsd that look as follow:

<xs:complexType name="ClassType">
<xs:sequence><xs:element name="IeplcHeader"><xs:complexType><xs:sequence><xs:element name="device-number" type="xs:integer" fixed="1"/></xs:sequence><xs:attribute name="version" type="xs:integer" use="required" fixed="0"/></xs:complexType></xs:element>

when i access with

doc.xpathEval('//xs:complexType/xs:sequence/xs:element[@name="IeplcHeader"]'):

tells me that cannot find the path.

while if i remove all the xs: as follow

<complexType name="ClassType"><sequence><element name="IeplcHeader"><complexType><sequence><element name="device-number" type="xs:integer" fixed="1"/></sequence><attribute name="version" type="xs:integer" use="required" fixed="0"/></complexType></element>

in this way it works

doc.xpathEval('//complexType/sequence/element[@name="IeplcHeader"]'):

Does anyone knows how can i get read of this problem fixing a prefix? righ now i am preparsing the file removing the xs: but it's an orrible solution and i really hope to be able to find a better solution.

(I did not try with py-dom-xpath yet and i do not know if may work even with the xs:)

thanks, ste

If you have to deal with xsd files, maybe also using them to validate xml files I suggest you to pass to lxml that has a good support for XMLSchema files.

example code:

from lxml import etree
from cStringIO import StringIOf = StringIO()f = StringIO('''\<xsd:schema xmlns:xsd="http://www.w3.org/2001/XMLSchema"><xsd:element name="a" type="AType"/><xsd:complexType name="AType"><xsd:sequence><xsd:element name="b" type="xsd:string" /></xsd:sequence></xsd:complexType></xsd:schema>
''')    xmlschema_doc = etree.parse(f)xmlschema_doc.xpath('xsd:element',namespaces={"xsd": "http://www.w3.org/2001/XMLSchema"})

results in:

[<Element {http://www.w3.org/2001/XMLSchema}element at 0x9a17f2c>]