Subject: XML to XML transformation From: Evyatar_Kafkafi <evyatar@xxxxxxxxx> Date: Wed, 22 Nov 2000 14:46:54 +0200 |
I'd like to use XSLT to transform an XML document to another XML document. The result should be very similar to the source, I may need to make only slight changes. To begin with, I tried to write an XSL stylesheet that creates a result document that is the same as the source. I found that I don't know how to do it. I searched all the XSL tutorials I could found on the web, and couldn't find an answer. The best thing I got was this: <?xml version="1.0" ?> <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"> <xsl:template match="/"> <xsl:apply-templates /> </xsl:template> <xsl:template match="*"> <xsl:text disable-output-escaping = "yes" ><</xsl:text> <xsl:value-of select="name()"/> <xsl:text disable-output-escaping = "yes" >></xsl:text> <xsl:apply-templates /> <xsl:text disable-output-escaping = "yes" ><</xsl:text> <xsl:value-of select="name()"/> <xsl:text disable-output-escaping = "yes" >/></xsl:text> </xsl:template> </xsl:stylesheet> There must be something simpler than this. Also, this worked for me with XSL Tester (of VBXML), but did not work with Xalan-J. Can anyone help me? Evyatar. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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