[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: transformation help
In XSLT 2.0 you can do this with <xsl:for-each-group select="*" group-starting-with="category"> <category name="{.}"> <xsl:copy-of select="current-group() except ."/> </category> </xsl:for-each-group> In 1.0, you can do: <xsl:for-each select="category"> <category name="{.}"> <xsl:copy-of select="following-sibling::item[generate-id(preceding-sibling::category[1]) = generate-id(current())]"/> </category> </xsl:for-each> In English, you are copying the following items whose imediately preceding category is the current category. Michael Kay > -----Original Message----- > From: Marcus B. Irven [mailto:marcus@xxxxxxxxxxxxxxx] > Sent: 14 June 2004 09:20 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: transformation help > > I have an xml doc in the form of > > <doc> > <category>cat 1</category> > <item>item 1</item> > <item>item 2</item> > <category>cat 1</category> > <item>item 3</item> > <item>item 4</item> > <item>item 5</item> > </doc> > > and i want to transform it into > > <doc> > <category name=3D"cat 1"> > <item>item 1</item> > <item>item 2</item> > </category> > <category name=3D"cat 2"> > <item>item 3</item> > <item>item 4</item> > <item>item 5</item> > </category> > </doc> > > how should i go about this? > > Thanks, > Marcus > > > --+------------------------------------------------------------------ > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > To unsubscribe, go to: http://lists.mulberrytech.com/xsl-list/ > or e-mail: <mailto:xsl-list-unsubscribe@xxxxxxxxxxxxxxxxxxxxxx> > --+-- > >
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