[XML-SIG] xslt/parameters
Matt Price
matt.price at utoronto.ca
Sat Aug 14 03:49:20 CEST 2004
Can someone out there tell me how I pass a parameter value to an xsl
stylesheet in python? Right now I have the following couple lines of
code, more or less stolen from somewhere since I'm still pretty much at
sea with xml:
styledoc = libxml2.parseFile(xsl)
style = libxslt.parseStylesheetDoc(styledoc)
doc = libxml2.parseDoc(risxSet)
result = style.applyStylesheet(doc, None)
htmlString = style.saveResultToString(result)
xsl is of course a variable which references a stylesheet. The
stylesheet has a parameter setting like this:
<xsl: param name="mainTarget">http://localhost/refdb-client/index.py</xsl:param>
I'd like to pass the parameter to the stylesheet in the above code.
Can this be done in a straightforward way? I get the impression I
should use the class libxslt.xpathParserContext(), but I really don't
understand how it's supposed to work! I much appreciate any pointers.
thanks,
matt
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Matt Price matt.price at utoronto.ca
History Department, University of Toronto
(416) 978-2094
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