Bug when copying processing-instruction() nodes via XSLT

Question

Hi,
Database version :
BANNER
Oracle Database 11g Express Edition Release 11.2.0.2.0 - Production
PL/SQL Release 11.2.0.2.0 - Production
CORE     11.2.0.2.0     Production
TNS for 32-bit Windows: Version 11.2.0.2.0 - Production
NLSRTL Version 11.2.0.2.0 - ProductionTest case (XSLT identity transform) :
SELECT XMLSerialize(document
         XMLTransform(
           XMLParse(document '<test><?abc?></test>')
         , XMLParse(document
'<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  <xsl:output method="xml"/>
  <xsl:template match="@*|node()">
    <xsl:copy>
      <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
  </xsl:template>
</xsl:stylesheet>')
       ) as result
FROM dual;produces :
RESULT
<?xml version="1.0" encoding="utf-8"?>
<?abc ?><test></test>
instead of the expected output, with the processing-instruction() node at the right place, in document order :
<?xml version = "1.0" encoding = "UTF-8"?>
<test><?abc ?></test>Any ideas or workarounds?
Thanks.

Answer 1

Yep, same on 11.2.0.3 :
SQL> select * from v$version;
BANNER
Oracle Database 11g Enterprise Edition Release 11.2.0.3.0 - Production
PL/SQL Release 11.2.0.3.0 - Production
CORE     11.2.0.3.0     Production
TNS for 32-bit Windows: Version 11.2.0.3.0 - Production
NLSRTL Version 11.2.0.3.0 - Production
SQL>
SQL> SELECT XMLSerialize(document
  2           XMLTransform(
  3             XMLParse(document '<test><?abc?></test>')
  4           , XMLParse(document
  5  '<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
  6    <xsl:output method="xml"/>
  7    <xsl:template match="@*|node()">
  8      <xsl:copy>
  9        <xsl:apply-templates select="@*|node()"/>
10      </xsl:copy>
11    </xsl:template>
12  </xsl:stylesheet>')
13           )
14         ) as result
15  FROM dual;
RESULT
<?xml version="1.0" encoding="utf-8"?>
<?abc ?><test></test>