Будет ли это работать для вас?
XSLT 1.0
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:fha="http://xmlns.oracle.com/F17CTOTALScorecard/F17CTOTALScorecard/F17CTOTALScorecard"
xmlns:client="http://xmlns.oracle.com/F17CTOTALScorecard/F17CTOTALScorecard/F17CTOTALScorecard">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<!-- identity transform -->
<xsl:template match="@*|node()">
<xsl:copy>
<xsl:apply-templates select="@*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="/fha:TOTALRequest-Response-Keys">
<soapenv:Envelope xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:f17="http://xmlns.oracle.com/F17CTOTALScorecard/F17CTOTALScorecard/F17CTOTALScorecard">
<soapenv:Body>
<xsl:apply-templates select="fha:TOTALRequest-Response/client:process"/>
</soapenv:Body>
</soapenv:Envelope>
</xsl:template>
<xsl:template match="client:*">
<xsl:element name="fha:{local-name()}">
<xsl:apply-templates/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
Что-то как это ?
public XDocument Serialize<T>(T source)
{
XDocument target = new XDocument();
XmlSerializer s = new XmlSerializer(typeof(T));
System.Xml.XmlWriter writer = target.CreateWriter();
s.Serialize(writer, source);
writer.Close();
return target;
}
public void Test1()
{
MyClass c = new MyClass() { SomeValue = "bar" };
XDocument doc = Serialize<MyClass>(c);
Console.WriteLine(doc.ToString());
}