Re: [xsl] Is a node a first/last child?

Subject: Re: [xsl] Is a node a first/last child?
From: Joerg Pietschmann <joerg.pietschmann@xxxxxx>
Date: Thu, 26 Jul 2001 10:07:33 +0200
> From: Nadia Karasawa <nadia@xxxxxxxxxxxxxxx>
> 
> 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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