The XML file name is specific but I need to build a dynamic path. I have tried using a variable to build the path but it didn't work:
<xsl:variable name="path">
...conditional c开发者_开发知识库ode
</xsl:variable> <xsl:value-of select="document('myXML.xml')/worksheets/$path"/>
2.0 solutions ok.
Evaluation of any dynamically-generated XPath expression is not supported by the XSLT 1.0 or XSLT 2.0 standards. It will be supported in XSLT 2.1.
If the dynamically-generated XPath expression is not too complex, the technique in this answer can be used successfully:
Retrieving XML node from a path specified in an attribute value of another node
You need an extension function, XPath 2.0 does not support dynamic compilation/evaluation. Saxon has saxon:evaluate
. Even if your processor does not support such function you might be able to implement it yourself as an extension function.
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