Hi, If that is the complete contents of the xsl, then what you should get back is all the text in the source file with the href attribute text from the CaseRef tags.
xslt has default templates that would match anything but CaseRef, but they only return text. Check the top answer here for an explanation: http://stackoverflow.com/questions/3360017/why-does-xslt-output-all-text-by-default /ch On Mon, Nov 24, 2014 at 5:58 AM, S.Gowtham <[email protected]> wrote: > > Hi all, > > > I invoked below transformation query to transform the xml with xslt. > > xdmp:xslt-invoke("example.xsl", $doc, (), ()) > > In example.xsl contains below contents. > > <xsl:template match="CaseRef"> > <xsl:variable xmlns:encoder="xalan://java.net.URLEncoder" > name="urlEncodedCit" > select="encoder:encode( substring( string(@href),2))"/> > <xsl:value-of select="$urlEncodedCit"/> > </xsl:template> > > But the marklogic query console returned file without any tags. It seems to > me that, the file is not transformed correctly. > > Can anyone help me to solve the issue. > > Thanks, > > Best Regards, > > $.Gowth@m > > _______________________________________________ > General mailing list > [email protected] > http://developer.marklogic.com/mailman/listinfo/general > _______________________________________________ General mailing list [email protected] http://developer.marklogic.com/mailman/listinfo/general
