RE: [xsl] Generating variable DOCTYPE

[Jarno.Elovirta@xxxxxxxxx]

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