从XML文档，XSLT和JAXB中删除元素

Question

This question is a follow up to my earlier question: Creating a valid XSD that is open using <all> and <any> elements

Given that I have a Java String containing an XML document of the following form:

<TRADE>
  <TIME>12:12</TIME>
  <MJELLO>12345</MJELLO>
  <OPTIONAL>12:12</OPTIONAL>
  <DATE>25-10-2011</DATE>
  <HELLO>hello should be ignored</HELLO>
</TRADE>

How can I use XSLT or similar (in Java by using JAXB) to remove all elements not contained in a set of elements. In the above example I am only interested in (TIME, OPTIONAL, DATE), so I would like to transform it into:

<TRADE>
  <TIME>12:12</TIME>
  <OPTIONAL>12:12</OPTIONAL>
  <DATE>25-10-2011</DATE>
</TRADE>

The order of the elements is not fixed.

Answer 1

This transformation:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

 <xsl:param name="pNames" select="'|TIME|OPTIONAL|DATE|'"/>
 <xsl:output omit-xml-declaration="yes" indent="yes"/>
 <xsl:strip-space elements="*"/>

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

 <xsl:template match="*/*">
  <xsl:if test="contains($pNames, concat('|', name(), '|'))">
   <xsl:call-template name="identity"/>
  </xsl:if>
 </xsl:template>
</xsl:stylesheet>

when applied on the provided XML document:

<TRADE>
    <TIME>12:12</TIME>
    <MJELLO>12345</MJELLO>
    <OPTIONAL>12:12</OPTIONAL>
    <DATE>25-10-2011</DATE>
    <HELLO>hello should be ignored</HELLO>
</TRADE>

produces the wanted, correct result:

<TRADE>
   <TIME>12:12</TIME>
   <OPTIONAL>12:12</OPTIONAL>
   <DATE>25-10-2011</DATE>
</TRADE>

Explanation :

1. The identity rule (template) copies every node "as-is".

2. The identity rule is overridden by a template matching any element that is not the top element of the document. Inside the template a check is made if the name of the matched element is one of the names specified in the external parameter $pNames in a pipe-delimited string of wanted names.

3. See the documentation of your XSLT processor on how to pass a parameter to a transformation -- this is implementation-dependent and differs from processor to processor.

Answer 2

I haven't tried yet, but maybe the javax.xml.tranform package can help:

http://download.oracle.com/javase/6/docs/api/javax/xml/transform/package-summary.html

Answer 3

JAXB & XSLT

JAXB integrates very cleanly with XSLT for an example see:

• How to get jaxb to Ignore certain data during unmarshalling

Your Other Question

Based on your previous question (see link below), the transform is really unnecessary as JAXB will just ignore attributes and elements that are not mapped to fields/properties in your domain object.

• Creating a valid XSD that is open using <all> and <any> elements