[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: applying xsl to xml to produce different xml
Hello, Thats easy. Try: <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:xtfile="http://www.jclark.com/xt/java/java.io.File" version="1.0" xmlns:xt="http://www.jclark.com/xt" extension-element-prefixes="xt"> <xsl:template match="books"/> <xsl:template match="authors"> <xsl:copy-of select="."> <xsl:apply-templates/> </xsl:copy-of> </xsl:template> <xsl:template match="author"> <xsl:copy-of select="."> <xsl:apply-templates/> </xsl:copy-of> </xsl:template> </xsl:stylesheet> Assume your file is called lib.xml and the stylsheet is lib.xsl, you'll want to try: xt lib.xml lib.xsl libout.xml -----Original Message----- From: owner-xsl-list@xxxxxxxxxxxxxxxx [mailto:owner-xsl-list@xxxxxxxxxxxxxxxx]On Behalf Of Clay_Rowland@xxxxxxxxxxx Sent: Thursday, December 02, 1999 12:07 PM To: xsl-list@xxxxxxxxxxxxxxxx Subject: applying xsl to xml to produce different xml I would like to apply a stylesheet to an xml doc and produce a new xml doc with only part of the original xml. for example: xml 1: <library> <books> <title>title 1</title> <title>title 2</title> </books> <authors> <author>author 1</author> <author>author 2</author> </authors> </library> into xml 2: <authors> <author>author 1</author> <author>author 2</author> </authors> i would like to return a new xml doc containg only the the node authors and its children nodes. i would like to retain the data in its tagged form. any suggestions? thanks. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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