Subject: RE: [xsl] Output character references for all but one character From: Pieter Reint Siegers Kort <pieter.siegers@xxxxxxxxxxx> Date: Mon, 21 Jun 2004 11:58:29 -0500 |
Hi Andrew, You seem to use xslt 2.0 but you instruct the processor to use version 1.0? HTH, <prs/> -----Original Message----- From: Andrew Welch [mailto:ajwelch@xxxxxxxxxxxxxxx] Sent: Monday, June 21, 2004 11:20 AM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: [xsl] Output character references for all but one character Consider the following xml: <node value="£">Hello World</node> I have the problem where I need characters outside the ascii range to be output as references, apart from the pound sign, which must remain as the actual character. The processor is Saxon 7.8 and is fixed. I can't use d-o-e as the pound sign is always an attribute value, and saxon 7.8 no longer supports d-o-e on attributes in favour of character maps. When I use the following stylesheet, I get a '?' character instead of the the pound, I guess because Saxon can't output the pound sign because it's not in the ascii range... <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:saxon="http://saxon.net.sf/" exclude-result-prefixes="saxon"> <xsl:output method="xhtml" encoding="ascii" omit-xml-declaration="yes" use-character-maps="style"/> <xsl:character-map name="style"> <xsl:output-character character="£" string="£"/> </xsl:character-map> <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> </xsl:stylesheet> Outputs: <node value="?">Hello World</node> I'm really after an exact copy of the input: <node value="£">Hello World</node> Any ideas? cheers andrew --+------------------------------------------------------------------ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/ or e-mail: <mailto:xsl-list-unsubscribe@xxxxxxxxxxxxxxxxxxxxxx> --+--
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