[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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