Subject: Re: [xsl] resolving relative path without base From: Kai Hackemesser <kaha@xxxxxx> Date: Wed, 30 May 2007 10:53:06 +0200 |
Ciao! Kai
Use StreamSource.setSystemId() to define the base URI.
Michael Kay
http://www.saxonica.com/
-----Original Message-----
From: Kai Hackemesser [mailto:kaha@xxxxxx] Sent: 29 May 2007 16:43
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
Subject: Re: [xsl] resolving relative path without base
Thanks, I tried that and it looks good.
But now I have a problem with relative links. My task is as follows: I have to create a xsl depending on some xml input. The resulting xsl file will import another xsl (depending on input from first xml file).
This is my code so far:
// loading preprocess XSL
StreamSource src = new
StreamSource(getClass().getResourceAsStream("/lrPreTransformHt
ml.xsl"));
// loading xml
StreamSource xml = new StreamSource(new ByteArrayInputStream(out.toByteArray()));
StreamResult result = new StreamResult(outRes); // create Transformer from factory and transform.
getTransformer(src).transform(xml,result);
// transform again and display
htmlPane.setText(transform(out.toByteArray(),outRes.toByteArray()));
The problem currently appearing is that the xsl file imported into the preprocess-created xsl stream also imports a xsl file - and since I now have only a ByteStream as Source, I lost the relativeness to that imported file. Is there another way than giving an absolute address in that import? Like setting a base url in FOP?
Kind regards, Kai
M. David Peterson schrieb:
On Tue, 29 May 2007 06:35:47 -0600, Kai Hackemesser<kaha@xxxxxx> wrote:
ame="xmlns:xsl">http://www.w3.org/1999/XSL/Transform</xsl:attribute>Has anybody seen an example how to create xsl output with xsl?<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0">
<xsl:template match="/">
<xsl:element name="xsl:stylesheet">
<xsl:attribute
<xsl:attribute name="version">1.0</xsl:attribute> <xsl:element name="xsl:template"> <xsl:attribute name="match">/</xsl:attribute> <xsl:element name="xsl:apply-templates"/> </xsl:element> </xsl:element> </xsl:template> </xsl:stylesheet>
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