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[xsl] With Saxon7.2, can an aliased namespace prefix be excluded from the output?
Subject: [xsl] With Saxon7.2, can an aliased namespace prefix be excluded from the output? From: "Timothy Larson" <Tdlarson@xxxxxxxxxxxxxxx> Date: Thu, 14 Nov 2002 13:28:19 -0500 |
With Saxon version 7.2, I am trying to apply a stylesheet to a stylesheet. Is there any way to exclude the aliased namespace prefix from the output? In the sample below I am trying to get rid of the output line: xmlns:x="http://www.w3.org/1999/XSL/Transform" Sample source stylesheet: <?xml version="1.0"?> <xsl:stylesheet version="2.0" xmlns:xsl = "http://www.w3.org/1999/XSL/Transform"> </xsl:stylesheet> Sample stylesheet to apply to the source stylesheet: <?xml version="1.0"?> <xsl:stylesheet version="2.0" xmlns:xsl = "http://www.w3.org/1999/XSL/Transform" xmlns:x = "http://www.w3.org/1999/XSL/TransformAlias" exclude-result-prefixes="x"> <xsl:namespace-alias stylesheet-prefix="x" result-prefix="xsl"/> <xsl:template match="/"> <xsl:copy> <x:stylesheet> <x:template m="/"><x:value-of select="test"/></x:template> </x:stylesheet> </xsl:copy> </xsl:template> </xsl:stylesheet> Output produced: <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:x="http://www.w3.org/1999/XSL/Transform" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template m="/"><xsl:value-of select="test"/></xsl:template> </xsl:stylesheet> Output disired: <?xml version="1.0" encoding="UTF-8"?> <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template m="/"><xsl:value-of select="test"/></xsl:template> </xsl:stylesheet> XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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