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Re: Is a node a first/last child?
- To: XSL List <xsl-list at lists dot mulberrytech dot com>
- Subject: Re: [xsl] Is a node a first/last child?
- From: Joerg Pietschmann <joerg dot pietschmann at zkb dot ch>
- Date: Thu, 26 Jul 2001 10:07:33 +0200
- Organization: ZKB
- Reply-To: xsl-list at lists dot mulberrytech dot com
> From: Nadia Karasawa <nadia@intraweb.com.br>
>
> I need to know if the current node is the first or the last child of his
> parent.
In order to clarify the answers you already got:
If list of the children is the context list, the most easy, fastest and
most readable solutions ought to be to use position(), as in the following
example:
<xsl:template match="parent_element">
<xsl:for-each select="node()">
<xsl:if test="position()=1">
First node of parent_element
</xsl:if>
<xsl:if test="position()=last()">
Last node of parent_element
</xsl:if>
</xsl:for-each>
<xsl:template>
If you have an arbitrary context, you'll have to stick to the
code Dimitre gave you (slightly optimized here :)
<xsl:template match="/">
<xsl:for-each select="key('random-key','some key value')>
<xsl:if test="preceding-sibling::node()">
First node of parent
</xsl:if>
<xsl:if test="following-sibling::node()">
Last node of parent
</xsl:if>
</xsl:for-each>
<xsl:template>
Note that preceding-sibling::node() also matches whitespace only text
nodes, so don't be surprised if the output omit some nodes which are
"clearly" the first/last child... Use a xsl:strip-whitespace or replace
node() in the code above by * if you deal with elements only.
Regards
J.Pietschmann
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