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Re: [xsl] How to strip all optional but empty elements from a XML

From: Michael Ludwig <mlu@---------.--->
To: xsl-list@-----.------------.---
Date: 8/4/2009 6:29:00 AM
Ben Stover schrieb:
> Assume I have a XML doc and a XSD schema file which defines mandatory
> and optional element fields.
>
> How can I now go recursively through the whole XML document

Recursion is free, as it is down by default.

> and find all elements which are optional (e.g. minoccurs=0)

That information is in XSD. I might be wrong, but I don't think it is
exposed in the XPath 2.0 Data Model with XSD annotations enabled. You
might get a list of the optional elements by (1) reading the schema,
(2) grepping the schema, or processing Saxon's normalized XSD
representation.

> but currently EMPTY.

"Empty" might need a better definition. I'll take it to mean the absence
of nodes on the child axis, without taking the attribute axis into
account.

> These empty, optional elements should be stripped off so that the
> resulting XML does not contain them any more.

Strippging is achieved by defining null template rules as follows.

<xsl:stylesheet version="1.0"
   xmlns:xsl="http://www.w3.org/1999/XSL/Transform">

   <xsl:template match="eins[not(node())]"/>
   <xsl:template match="zwei[not(node())]"/>
   <xsl:template match="drei[not(node())]"/>

   <xsl:template match="@*|node()"><!-- identity template -->
     <xsl:copy><xsl:apply-templates select="@*|node()"/></xsl:copy>
   </xsl:template>

</xsl:stylesheet>

Michael Ludwig

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