get node name from namespace - xslt

Question

this is my xml
<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:end="http://endpoint.ggg.com/">
<soapenv:Header/>
<soapenv:Body>
<end:onlineExpressRemit>
<channelCode>NBPS</channelCode>
</end:onlineExpressRemit>
</soapenv:Body>
</soapenv:Envelope>
and this is my xslt
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet version="1.0" xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:end="http://endpoint.ggg.com/" >
<xsl:template match="/">
<xsl:value-of name="ggg2" select="/soapenv:Envelope/soapenv:Body/*[namespace-uri()]"/>
</xsl:template>
</xsl:stylesheet>
my expected output is display only
onlineExpressRemit
i am very new for xslt, i try name(), local-name() alot others method, however no luck, i cant retrieve only node-name, any help will do, thank you!

Answer 1

You can try using local-name(soapenv:Envelope/soapenv:Body/*), for example, the following XSL transformation :
<xsl:stylesheet version="1.0" xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:end="http://endpoint.ggg.com/" >
<xsl:output method="text" omit-xml-declaration="yes"/>
<xsl:template match="/">
<xsl:element name="root">
<xsl:value-of select="local-name(soapenv:Envelope/soapenv:Body/*)"/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
when applied to your XML sample, yield the following output :
<root>onlineExpressRemit</root>
Note: I added <root> element just because in online XSLT processor I used for testing, there was error when root element is absent. Usually, setting <xsl:output> is enough to be able to output a non-XML result : XSLT: Transforming into non-xml content?