Subject: RE: [xsl] Generating variable DOCTYPE From: Jarno.Elovirta@xxxxxxxxx Date: Thu, 16 Jan 2003 11:00:07 +0200 |
Hi, > 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) Yes. > and without extension functions. Is this > possible somehow ? There is no XSLT 1.0 solution; use your XSLT engine's API to control the serialization. Cheers, Jarno - De/Vision: Endlose Traume XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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