python - 将评估的xpath传递给扩展

我正在使用lxml python lib。

假设我们有产品xml，例如：

<product id='123' />

并要应用xsl模板：

<xsl:template match="product">
    <ssi:include virtual="/ssi/reviews/{@id}"/>
</xsl:template>

ssi：include是一个简单的lxml扩展，可将nginx ssi指令作为注释插入HTML代码中。问题是评估@id并将属性传递为virtual =“ / ssi / include / 123”。有办法吗？
我已经找到了一种解决方案并在使用它：

import lxml.etree
import re
from copy import deepcopy

ns = '{ssi}'

# ssi extensions
class SsiExtElement(lxml.etree.XSLTExtension):
    def execute(self, context, self_node, input_node, output_parent):
        _, tag = self_node.tag.split('}')
        tmp = lxml.etree.Element('tmp')
        for k, v in self_node.attrib.items():
            if re.search('\{(.*)\}', v): #here we search {xpath} values to evaluate
                elem = deepcopy(input_node)
                matches = re.findall('\{(.*)\}', v)
                for match in matches:
                    v = v.replace('{%s}' % match, elem.xpath(match)[0])
            tmp.set(k, v)
        self.process_children(context, output_parent=tmp)
        attrs = ' '.join(u'%s="%s"' % (k, v) for k, v in tmp.attrib.items())
        ssi = lxml.etree.Comment(u'#%s %s' % (tag, attrs))
        output_parent.append(ssi)
        for node in tmp:
        output_parent.append(node)
        if (self_node.tag.replace(ns,'') in ('if', 'else', 'elif')
            and self_node.getnext().tag.replace(ns, '') not in ('else', 'elif')):
        output_parent.append(lxml.etree.Comment(u'#endif'))

最佳答案

尝试使用xsl：attribute

<xsl:template match="product">
    <ssi:include>
       <xsl:attribute name="virtual">
          <xsl:value-of select="concat('/ssi/reviews/',@id)"/>
       <xsl:attibute>
    </ssi:include>
</xsl:template>