Re: [xsl] how to store node in variable?

Subject: Re: [xsl] how to store node in variable?
From: John <john-xsl-list@xxxxxxxx>
Date: Thu, 21 Oct 2004 13:59:56 -0700
Thanks. Unfortunately that didn't really help me, but probably because I didn't describe my problem well. But, I do seem to have found an alternative. Not sure if this will make sense out of context:

 <xsl:choose>
   <xsl:when test="some condition">
     <xsl:for-each select="...">
       <xsl:call-template name="whatever" />
     </xsl:for-each>
   </xsl:when>
   <xsl:otherwise>
     <xsl:call-template name="whatever" />
   </xsl:otherwise>
 </xsl:choose>

I don't know why I didn't think of this in the first place.

G. Ken Holman wrote:

At 2004-10-21 12:39 -0700, John wrote:

How do I store a reference to a node in a variable


Only in the select= attribute.

or what am I doing wrong? Here is what I have (maybe oversimplified) :

<xsl:variable name="me">
  <xsl:choose>
    <xsl:when test="<some condition>">
      <xsl:copy-of select=".." />
    </xsl:when>
    <xsl:otherwise>
      <xsl:copy-of select="." />
    </xsl:otherwise>
  </xsl:choose>
</xsl:variable> <xsl:call-template name="<some template>">
  <xsl:with-param name="myid" select="$me/@id" />
</xsl:call-template>

The with-param is always throwing an error, expression should result in a node-set.


Right ... because it is a result tree fragment, not a node set.

If I add msxsl:node-set around my selects in the variable definition I get cannot convert result tree fragment to node-set.


Not sure why, but you'd get a copy of the node, not the node itself, so you would lose the context, so that wouldn't help (and it wouldn't be pure XSLT).

Alternatively, is there any way to explicitly set the context node without using for-each?


I would use:

    <xsl:variable name="condition" select="<some condition>"/>
    <xsl:variable name="me" select="../self::*[$condition] |
                                    self::*[not($condition)]"/>

Only one of the union would be in the result because the condition is true exactly once.

I hope this helps.

........................ Ken

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