Subject: RE: [xsl] Removing elements based on contents From: "Michael Kay" <mike@xxxxxxxxxxxx> Date: Sat, 24 Jun 2006 15:40:50 +0100 |
In XSLT you don't remove the elements you want to lose, you copy the elements you want to keep. So: <xsl:template match="Root"> <xsl:copy> <xsl:copy-of select="Story[.='End Here']/following-sibling::*"/> </xsl:copy> </xsl:template> Michael Kay http://www.saxonica.com/ > -----Original Message----- > From: Chad Chelius [mailto:cchelius@xxxxxxxxxxxxxxx] > Sent: 24 June 2006 13:04 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: [xsl] Removing elements based on contents > > I have an XML file that looks something like this: > > <Root> > <Story>Content</Story> > <Story>More Content</Story> > <Story>End Here</Story> > <Story>Good stuff that I want</Story> > </Root> > > My question is: Is there a way using XSLT to remove all > elements up to and including the one who's content contains > <Story>End Here</ > Story> and leave the rest intact? Basically everything from the top > of the XML file down to and including that tag is junk that I > don't want to include in the file but the rest of it I want > to keep. I don't think XSLT traverses I file in that way > though. Does anyone have any ideas?
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