[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] XSL to XSL stylesheet: namespace question
Hello xsl-list, could anyone give me an insight on how to make correct transform from xsl to xsl. Here is an example (the bad thing is that it makes ns0:stylesheet xmlns:ns0="http://www.w3.org/1999/XSL/Transform" and thus xsl goes as undeclared ns): source xsl: <?xml version="1.0" encoding="UTF-8"?> <?xml-stylesheet type="text/xsl" href="design.xsl"?> <adt:root xmlns:adt="http://somewhere.domain/ADT" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <html> Some here </html> </adt:root> <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:adt="http://somewhere.domain/ADT"> <xsl:output method="xml" indent="yes" version="1.0" encoding="UTF-8"/> <xsl:template match="/"> <xsl:apply-templates/> </xsl:template> <xsl:template match="adt:root"> <xsl:element name="xsl:stylesheet" namespace="http://www.w3.org/1999/XSL/Transform"> <xsl:attribute name="version">1.0</xsl:attribute> <xsl:element name="xsl:output"> <xsl:attribute name="method">xml</xsl:attribute> <xsl:attribute name="version">1.0</xsl:attribute> <xsl:attribute name="encoding">UTF-8</xsl:attribute> <xsl:attribute name="indent">yes</xsl:attribute> </xsl:element> <xsl:apply-templates/> </xsl:element> </xsl:template> <xsl:template match="*|@*"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> </xsl:stylesheet> -- Best regards, viewga mailto:viewga@xxxxxxxxxxxx XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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