[XML-SIG] xslt/parameters
Daniel Veillard
veillard at redhat.com
Sat Aug 14 11:17:12 CEST 2004
On Fri, Aug 13, 2004 at 09:49:20PM -0400, Matt Price wrote:
> 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,
You're using libxml2/libxslt in that context, better as for help
in the right channel
http://xmlsoft.org/XSLT/bugs.html
the parameter to the transformation are passed as a dictionnary
to applyStylesheet(), instead of passing None, pass the dictionary
containing the (name, value) pairs for all parameters.
Daniel
--
Daniel Veillard | Red Hat Desktop team http://redhat.com/
veillard at redhat.com | libxml GNOME XML XSLT toolkit http://xmlsoft.org/
http://veillard.com/ | Rpmfind RPM search engine http://rpmfind.net/
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