[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: Splitting attributes
Endre,
Also, if your IDREFS attribute actually happens to point to IDs (which they should, as IDREFs), and if the IDs are declared in a DTD available to the transformation, you can exploit a feature of the id() function. id() will work with IDREFS values, so if you said <xsl:template match="Class"> <xsl:for-each select="id(@subtypes)"> ... </xsl:for-each> </xsl:template> The for-each will iterate over the nodes that carry the ID attributes corresponding to your IDREFS value. Even if you don't actually want to process those nodes, it's a cheap way of getting to the IDREFS values separately (since of course they'll also be the IDs on the nodes retrieved by the function). See XPath 4.1 for the lowdown on id(). Cheers, Wendell At 02:10 AM 11/18/2002, Oleg wrote: Endre Magyari wrote:In my source XML I have elements with IDREFS attributes, like: <Class subtypes="id1 id23 id56 id34"> My desire is to generate an element for each idref in part.(Split the attribute to idrefs) Any idea how to do this? I haven't seen any splitting functions in the XPATH documentation. Mostly, I would like an <xsl:for-each>-like behaviour to iterate on the set of the strings resulted from splitting the attribute value. ====================================================================== Wendell Piez mailto:wapiez@xxxxxxxxxxxxxxxx Mulberry Technologies, Inc. http://www.mulberrytech.com 17 West Jefferson Street Direct Phone: 301/315-9635 Suite 207 Phone: 301/315-9631 Rockville, MD 20850 Fax: 301/315-8285 ---------------------------------------------------------------------- Mulberry Technologies: A Consultancy Specializing in SGML and XML ====================================================================== XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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