[xsl] Obtain XSL content

[Márcio Fernando Keller <marcio@xxxxxxxxxxxxx>]

Hi!

What is the best way to obtain the content of a xsl file. Ex:

<?xml version="1.0"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform";
version="1.0">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="*|@*">
    <xsl:copy>
        <xsl:apply-templates select="@*|node()"/>
    </xsl:copy>
</xsl:template>

<xsl:template match="/Orders/SalesOrder/Customer/Gender">
    <xsl:copy>
        <xsl:choose>
            <xsl:when test="/Orders/SalesOrder/Customer/Gender='F'">
                <xsl:text>Female</xsl:text>
            </xsl:when>
        <xsl:when test="/Orders/SalesOrder/Customer/Gender='M'">
            <xsl:text>Male</xsl:text>
        </xsl:when>
        </xsl:choose>
    </xsl:copy>
</xsl:template>
</xsl:stylesheet>

I need to know that exist a template  for
"/Orders/SalesOrder/Customer/Gender" and tests with choose, ....
I think that i can use parser classes(ex xalan), xpath or process it like
another xml file, but what the best method for this?
Examples?

Thanks


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