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[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Obtain XSL content
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
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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