Subject: RE: [xsl] Convert String to (node-set) From: mxmodi@xxxxxxxxxxxxxxx Date: Mon, 5 Feb 2001 10:10:22 -0600 |
I tried using the saxon"evaluate function. However, I got an error in expression when I processed the XSL using Saxon 5.4. What am I missing? Snippet of XSL code: <xsl:stylesheet version="1.0" xmlns:xsl ="http://www.w3.org/1999/XSL/Transform" xmlns:saxon="http://icl.com/saxon"> <xsl:variable name="node" select="document('BER.xml')/CHANNEL/EVENT/NODE" /> <xsl:variable name="chandoc" select="document('TSInsertBER.xml')" /> <xsl:template match="/"> <xsl:for-each select="$node//Node_path"> <xsl:text>The Value retrieved from doc 2 is: </xsl:text><xsl:value-of select="saxon:evaluate($chandoc/.)" /> </xsl:for-each> </xsl:template> <xsl:stylesheet> "Michael Kay" <mhkay@xxxxxxxxxxxx> To: <xsl-list@xxxxxxxxxxxxxxxxxxxxxx> Sent by: cc: owner-xsl-list@xxxxxxxxxxxx Subject: RE: [xsl] Convert String to (node-set) rytech.com 02/03/2001 12:43 PM Please respond to xsl-list > I am trying to parse two separate XML documents in one XSL > stylesheet. This > I can do!. My Q is, I am extracting the value of a node from > XML doc. 1 and > then I want to pass that as XPATH for parsing XML doc 2. There is no standard way in XSLT of evaluating an XPath expression constructed as a string or read from another document. You can do it easily in Saxon using saxon:evaluate(), and in MSXML3 with JavaScript extension functions using selectNodes(). Mike Kay 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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