XML/XSLT with Python/lxml

xslt.py

#!/usr/bin/python

# transform any XML with a XSLT
# you can pass additional parameters for your stylesheet

def xsl_transformation(xslfile, xmlfile = None, xmlstring = None, params={}):
    from lxml import etree
    import StringIO
    xslt_tree = etree.XML(utils.readfile(xslfile))
    transform = etree.XSLT(xslt_tree)
    xml_contents = xmlstring
    if not xml_contents:
        if xmlfile:
            xml_contents = utils.readfile(xmlfile)
        else:
            xml_contents = '<?xml version="1.0"?>\n<foo>A</foo>\n'
    f = StringIO.StringIO(xml_contents)
 
    doc = etree.parse(f)

    f.close()
    transform = etree.XSLT(xslt_tree)
    result = transform(doc, **params)
   
    return result

Thanks a lot! You forgot an f in the name of the function xsl_transormation. ;-)

Will this supports XSLT2.0

lxml only supports XSLT 1.0

Is there an alternative to using lxml? I have a usecase where I cannot install lxml as part of the package.