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Re: [xsl] Saxon 9.4 Transformed to (newline) Problem

Subject: Re: [xsl] Saxon 9.4 <bold></bold> Transformed to (newline)</bold> Problem
From: Michael Kay <mike@xxxxxxxxxxxx>
Date: Tue, 15 Jan 2013 17:03:51 +0000

There are several ways you could fix this problem.

First, you could switch indentation off entirely (indent="no").

Secondly, you could use schema-awareness on the output side (that is, validate the output against a schema). If you do this, Saxon will not add whitespace within an element that has a mixed-content content model. Validation of course needs Saxon-EE.

Thirdly, you could use the suppress-indentation xsl:output parameter. This was introduced as a Saxon extension (saxon:suppress-indentation) and has found its way into the XSLT 3.0 specification. Either way, you will need Saxon-PE or higher.

Note that the fact that the bold element is empty has nothing to do with it; the indentation will occur for any start tag unless it is suppressed.

I assume that you're not complaining about the translation of <bold></bold> to <bold/>?

Michael Kay

On 15/01/2013 14:14, Raymond Lillibridge wrote:
List members,

Due to some batch processing, some of my input XML may have empty elements.

Here is some sample XML:
<para> Here is some text inside a para tag. <bold></bold> Note that the 'bold' element before the word Note is empty.  I would like it to stay that way without the insertion of a newline.</para> </level1>

When I transform this XML, using Saxon 9.4, the <bold></bold> element is getting converted similar to the following: <level1> <para> Here is some text inside a para tag. <bold/> Note that the 'bold' element before the word Note is empty. I would like it to stay that way.</para> </level1>

The Problem: Due to further batch processing needs, I do not want the insertion of a newline before the <bold/> element, which is being created after running an XSLT transformation on the sample XML above. (XMLSpy does not insert the newline, by the way, but I want to use Saxon for my transformation.) In my xsl file I do not have an explicit template match for the 'bold' element.

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="2.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"

<xsl:strip-space elements="*" />
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>

<!-- To Get:  {$InputDocPath} -->
<xsl:include href="./MCC_LIB.xsl"/>

<xsl:template match="/">
	<xsl:result-document href="{$InputDocPath}/Book_ALL.xml">
		<xsl:apply-templates select="node()" />
<xsl:template match="book">
		<xsl:element name="book">
			<xsl:element name="bookinfo">
				<xsl:element name="title"></xsl:element>
				<xsl:element name="subtitle"></xsl:element>
			<xsl:apply-templates select="node()"/>	

<xsl:template match="level1|level2|level3|level4|level5|level6"> <xsl:copy-of select="./node()" copy-namespaces="no" /> </xsl:template>

<!-- CATCH-ALL ==================================================== --> <xsl:template match="@*|node()"> <xsl:copy-of select="./node()" copy-namespaces="no" /> </xsl:template> </xsl:stylesheet>

Looking in the Saxon documentation, I was not able to find a switch to control the transformation behavior that changes the <bold></bold> to (newline)<bold/>.

I'd rather not use XMLSpy since all my other batch transformations are using Saxon.
If there is a configuration switch for Saxon, could someone direct me where I may learn about it.

Or, would it be more practical to write a template to remove the "empty" <bold></bold> element, or better yet, remove all empty elements? I don't know how this would be written, and would appreciate any insights someone may offer.

Kind regards,

Raymond Lillibridge
Sr. Software Engineer
Municipal Code Corporation | Facebook | Twitter

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