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(_at_)agitraining(_dot_)com]
Sent: 24 June 2006 13:04
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
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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