Subject: Re: How do we generate DTD from XSL? From: Duane Nickull <webmaster@xxxxxxxxxxxxxxxxx> Date: Fri, 23 Apr 1999 12:33:26 -0700 |
Yvon Sauvageau wrote: > > Hi, > > I'm using XSL to convert from one XML format to another XML format. I > figured out how to generate the <?xml version="1.0"?> tag: I just have to > write <xsl:pi name="xml">version="1.0"</xsl:pi>. But I don't see how to > generate the <!DOCTYPE mydoc SYSTEM "mydoc.dtd"> tag. Is there any way to do > it? Have you tried using entity references like this: <xsl:pi name="xml">version="1.0"</xsl:pi> <!DOCTYPE mydoc SYSTEM "<xsl:value-of select="filename_from_your_xml_document.dtd"/>< <br/> Then place the rest of your document here Depending on what you are using to view the result (IE5.0), you may also need to declare other entities. Once you have delcared an entity in the input tree, you may freely reference it in the output tree since any validating parser will store that entity's value. Duane Nickull > > Thanks, > > Yvon Sauvageau > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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