Subject: Re: node-set() function in MSXML? From: "Steve Muench" <smuench@xxxxxxxxxxxxx> Date: Tue, 1 Aug 2000 16:49:48 -0700 |
| | Does MSXML implement a node-set() function (in a standard way)? | Not sure what you mean by "in a standard way", since the node-set() function is not part of the XSLT 1.0 Recommendation, but my experience playing with MSXSL3 (retested just now on the July 2000) is that MSXSL3 does not require a node-set() function because it treats Result Tree Fragments as node-sets Try the following with MSXSL3 (using any convenient XML file as input): <test xsl:version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:variable name="x"> <a> <b>one</b> <b>two</b> </a> </xsl:variable> <xsl:for-each select="$x/a/b"> <c><xsl:value-of select="."/></c> </xsl:for-each> </test> You'll get: <?xml version="1.0" encoding="UTF-16"?> <test><c>one</c><c>two</c></test> as output. Late-model versions of Saxon, OracleXSL, XT, and Xalan complain about "cannot convert result tree fragment", so require an extension function to achieve this. ______________________________________________________________ Steve Muench, Lead XML Evangelist & Consulting Product Manager BC4J & XSQL Servlet Development Teams, Oracle Rep to XSL WG Author "Building Oracle XML Applications", O'Reilly http://www.oreilly.com/catalog/orxmlapp/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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