Re: xpath - how to return all nodes but the node match
Hi Scott, > In order to "delete" a node, I'm trying to write an xpath that, when > given an id, will return the entire tree with the exception of the > matching node for the id. I've tried many possibilities but can't > seem to get the thing to work -- many times it just skips the match > all together and returns the whole tree. Since the heirarchical > structure is arbitrary, I cannot just pinpoint the exact location of > the node to be removed -- I suppose I could with two stylesheets, > but I have a feeling there is a more elegant solution with a single > xpath. > > Can someone please help me with the xpath that would return the > whole tree, minus the node? Trying to select everything bar the node you want to delete, and then copying those nodes, won't work because you'll lose the structure. The XSLT processor will create a deep copy of every node that you select, and you'll end up with a a copy of the entire file for the root node, another copy of the entire file for the document element, a copy of the first element and all its descendants for the first element child of the document element and so on. The best approach for a filtering stylesheet such as this is to have an identity template that moves through the file recursively copying each node except for the one that you want to omit. Start with the basic identity template: <xsl:template match="@*|node()"> <xsl:copy> <xsl:apply-templates select="@*|node()" /> </xsl:copy> </xsl:template> Then modify it so that the copying of the matched node doesn't happen if it has an id attribute whose value is the value of the $delete_id parameter: <xsl:template match="@*|node()"> <xsl:if test="not(@id = $delete_id)"> <xsl:copy> <xsl:apply-templates select="@*|node()" /> </xsl:copy> </xsl:if> </xsl:template> Cheers, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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