[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: Re: file manipulation with recursion
Thanks for all your help. I have a little twist to this file however. Under each category node there could be zero, one or many product nodes. I want to just display the id and name of the product node as well, and they can appear under any category node. The initial file format would be of the following... hierarchy> <category> <id>0</id> <level>-1</level> <name>Cat1</name> <releaseLevel>Live</releaseLevel> <date>2002-02-25 12:29:46</date> <category> <id>13abc</id> <level>1</level> <name>Cat2</name> <releaseLevel>Live</releaseLevel> <date>2002-01-07 14:02:41</date> <category> <id>X12345</id> <level>2</level> <name>Cat3</name> <releaseLevel>Live</releaseLevel> <date>2002-07-11 14:52:06</date> <product> <id>abc123</id> <type>new</type> <name>myProduct</name> <releaseLevel>Live</releaseLevel> <date>2002-07-11 14:52:06</date> </product> <product> <id>abc987</id> <type>new</type> <name>myProduct3</name> <releaseLevel>Live</releaseLevel> <date>2002-07-11 14:52:06</date> </product> </category> <product> <id>asd123</id> <type>old</type> <name>myProduct2</name> <releaseLevel>Live</releaseLevel> <date>2002-07-11 14:52:06</date> </product> </category> <category> <id>13abc</id> <level>1</level> <name>Cat2</name> <releaseLevel>Live</releaseLevel> <date>2002-01-07 14:02:41</date> </category> </category> </hierarchy> Thanks again for your help! Rick -----Original Message----- From: Dimitre Novatchev [mailto:dnovatchev@xxxxxxxxx] Sent: Tuesday, July 23, 2002 8:41 AM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: file manipulation with recursion --- Holmberg Rick-ra0119 <Rick dot Holmberg at motorola dot com> wrote: > Hi all, > I am new to XML/XSLT and have a file that I need to generate a > smaller version of. The file is of the format: > > <hierarchy> > <category> > <id>0</id> > <level>-1</level> > <name>Cat1</name> > <releaseLevel>Live</releaseLevel> > <date>2002-02-25 12:29:46</date> > <category> > <id>13abc</id> > <level>1</level> > <name>Cat2</name> > <releaseLevel>Live</releaseLevel> > <date>2002-01-07 14:02:41</date> > <category> > <id>X12345</id> > <level>2</level> > <name>Cat3</name> > <releaseLevel>Live</releaseLevel> > <date>2002-07-11 14:52:06</date> > </category> > </category> > </category> > </hierarchy> > > I need to have the output file be of the format > <hierarchy> > <category> > <id>0</id> > <name>Cat1</name> > <category> > <id>13abc</id> > <name>Cat2</name> > <category> > <id>X12345</id> > <name>Cat3</name> > </category> > </category> > </category> > </hierarchy> Hi Rick, The following transformation produces the desired results: <xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" > <xsl:output omit-xml-declaration="yes" indent="yes"/> <xsl:strip-space elements="*"/> <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()"/> </xsl:copy> </xsl:template> <xsl:template match="category/*[not(self::id or self::name or self::category ) ]"/> </xsl:stylesheet> This is just the identity rule plus a rule to eliminate any children of a "category", which are not "id" or "name" or "category" themselves. And when applied to your source xml document, the result is exactly as desired: <hierarchy> <category> <id>0</id> <name>Cat1</name> <category> <id>13abc</id> <name>Cat2</name> <category> <id>X12345</id> <name>Cat3</name> </category> </category> </category> </hierarchy> Hope this helped. ===== Cheers, Dimitre Novatchev. http://fxsl.sourceforge.net/ -- the home of FXSL __________________________________________________ Do You Yahoo!? Yahoo! Health - Feel better, live better http://health.yahoo.com 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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