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Hi, I have a XML document which resembles, <?xml version="1.0" encoding="ISO-8859-1"?> <CONTENT TYPE="PR"> <HEADER>Header</HEADER> <TITLE>Title</TITLE> <SDATE>09-Feb-2000</SDATE> <SUMMARY> <STEXT>Description</STEXT> <IMGSRC SIZE="1">images/dwc_m.gif</IMGSRC> </SUMMARY> <SUBTITLES> <SUBTITLE ID="1"> <KEYWORDS>Keyword</KEYWORDS> <HEADING>The Event</HEADING> <LINKTXT TXTMAXLEN="">The Event</LINKTXT> <BODY><![CDATA[SomeData]]></BODY> <IMGSRC SIZE="">images/dwc_m.gif</IMGSRC> <IMGCAPT>Image caption</IMGCAPT> </SUBTITLE> </SUBTITLES> <RELATEDLINKS> <RLINK ID="1" TARGET="POPUP"> <LTEXT MAXLEN="">Some heading</LTEXT> <URL>http://www.something.com</URL> </RLINK> </RELATEDLINKS> <FOOTER>footer</FOOTER> </CONTENT> I need to tranform this XML into encoding type "ISO-8859-6" (Arabic). I'm using the Oracle-XML Parser via Java to perform the transformation. So I used this XSLT, <?xml version="1.0" ?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" > <xsl:output method="xml" encoding="ISO-8859-6" cdata-section-elements="BODY" /> <xsl:apply-templates select="/" /> <xsl:template match="/"> <xsl:copy-of select="/" /> </xsl:template> </xsl:stylesheet> This does the job fine. Now I want to make my XSLT more generic so I don't have to specifically say that the "<BODY>" is a CDATA section. Basically I want an exact copy of the original document whatever the original document may be. I tried, <?xml version="1.0" ?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0" > <xsl:output method="xml" encoding="ISO-8859-6" cdata-section-elements="BODY" /> <xsl:apply-templates select="/" /> <xsl:template match="/"> <xsl:copy-of select="document()" /> </xsl:template> </xsl:stylesheet> but this doesn't work. Much appreciated if you could give me a hint on how to do this. And also any pointers on whether I'm doing something completely wrong. Thanks and regards, Mayura XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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