Subject: RE: [xsl] Generating variable DOCTYPE From: Edward.Middleton@xxxxxxxxxxx Date: Thu, 16 Jan 2003 18:06:42 +0900 |
Like this <?xml version="1.0"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:param name="DTDLocation"/> <xsl:template match="/"> <xsl:text disable-output-escaping="yes"><![CDATA[ <!DOCTYPE svg PUBLIC "-//W3C//DTD SVG 20001102//EN" "]]></xsl:text> <xsl:value-of select="$DTDLocation"/> <xsl:text disable-output-escaping="yes"><![CDATA[">]]></xsl:text> </xsl:template> </xsl:stylesheet> Edward Middleton -----Original Message----- From: Cams Ismael [mailto:Ismael.Cams@xxxxxxxxxxxxxxx] Sent: Thursday, January 16, 2003 5:35 PM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: [xsl] Generating variable DOCTYPE Hello, I am trying to transform an xml file into another xml file by means of a stylesheet. This stylesheet takes as input parameter the location of the DTD belonging to the generated xml file. What I would like to do is: <xsl:param name="DTDLocation"/> <xsl:output method="xml" encoding="UTF-8" doctype-system="{$DTDLocation}"/> This is possible in XSLT 1.1, but not in XSLT 1.0. Another way is to write an extension function that writes the DOCTYPE to the output. However I prefer a solution in XSLT 1.0 (if I am correct XSLT 1.1 became never a Recommendation) and without extension functions. Is this possible somehow ? Thanks in advance. Kind regards, Ismaël XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] Generating variable DOCTY, Mike Brown | Thread | RE: [xsl] Generating variable DOCTY, Cams Ismael |
RE: [xsl] Generating variable DOCTY, Jarno . Elovirta | Date | RE: [xsl] Generating variable DOCTY, Cams Ismael |
Month |