[XSL-LIST Mailing List Archive Home]
[By Thread]
[By Date]
RE: [xsl] Removing elements based on contents
Subject: RE: [xsl] Removing elements based on contents From: cknell@xxxxxxxxxx Date: Sat, 24 Jun 2006 10:41:17 -0400 |
Use this template for <Story> to achieve what you want. <xsl:template match="Story"> <xsl:choose> <xsl:when test="text() = 'End Here'"></xsl:when> <xsl:when test="following-sibling::Story[text() = 'End Here']"></xsl:when> <xsl:otherwise><xsl:copy-of select="." /></xsl:otherwise> </xsl:choose> </xsl:template> -- Charles Knell cknell@xxxxxxxxxx - email -----Original Message----- From: Chad Chelius <cchelius@xxxxxxxxxxxxxxx> Sent: Sat, 24 Jun 2006 08:03:53 -0400 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?
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
RE: [xsl] Removing elements based o, Michael Kay | Thread | [xsl] Solution: for Recursively mer, Laky Tang |
RE: [xsl] Removing elements based o, Michael Kay | Date | Re: [xsl] [FXSL] Using document(''), Florent Georges |
Month |