Subject: AW: [xsl] how to "save" a context? From: "Braumüller, Hans" <H.Braumueller@xxxxxxxxxxxx> Date: Thu, 16 Jan 2003 13:38:39 +0100 |
Ho Dimitri, i tried to do it with msxml3 and 4 <?xml version="1.0" encoding="ISO-8859-1" ?> <?xml-stylesheet type="text/xsl" href="test.xsl"?> <root> <test>xpath</test> </root> and <xsl:stylesheet version='1.0' xmlns:xsl='http://www.w3.org/1999/XSL/Transform'> <xsl:output method="text" /> ..... (nothing changed) <xsl:value-of select="$theResult"/> and get as result ÿþ Never mind, which xsml-source i use, i get the same. What i do wrong ? Thanks, Hans Braumüller "S Woodside" <sbwoodside@xxxxxxxxx> wrote in message news:1D578741-27F8-11D7-AD3E-000393414368@xxxxxxxxxxxx > Gah! It's huge! > > Seriously it looks like this will generate a string that, when assigned > to a variable, and selected upon, will act as the correct XPath. Is > that right? Yes, actually this snippet processes a nodeset of nodes and generates a set of strings (XPath expressions) for each of them. You can simplify it a bit to work only on a single node (get rid of the outermost xsl:for-each). > > It has: <xsl:value-of select="msxsl:node-set($theResult)"/> > > Does it require MSxml? Can I turn $theResult (which is a string, > right?) into an XPath expression I can use in a select statement, > without needing MSXML? Actually, the xxx:node-set extension function is not required. Just get rid of it and the msxsl namespace, too. ===== Cheers, Dimitre Novatchev. http://fxsl.sourceforge.net/ -- the home of FXSL __________________________________________________ Do you Yahoo!? Yahoo! Mail Plus - Powerful. Affordable. Sign up now. http://mailplus.yahoo.com XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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