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MyASPN >> Mail Archive >> xsl-list
xsl-list
Re: [xsl] XPath question
by Jeni Tennison other posts by this author
Aug 24 2001 10:00AM messages near this date
[xsl] XPath question | [xsl] Formatting bold and italics within a paragraph
Hi Adam,

>  Now what is all the all the attrib nodes from $rules whose name AND parents
>  state attribute are not present together in $attribs.

I don't think it's possible to do in a single XPath. You can do:

  <xsl:for-each select="$rules/attrib"> 
    <xsl:if test="not($attribs[@name = current()/@name and
                               @state = current()/../@state])"> 
      ...
    </xsl:if> 
  </xsl:for-each> 

If the attrib elements in $attribs were held somewhere other than in a
variable, then you could construct a key for them:

<xsl:key name="attribs" match="attrib"
         use="concat(@name, ' ', @state)" /> 

And then you could do:

  $rules/attrib[not(key('attribs', concat(@name, ' ', ../@state)))]

I hope that helps,

Jeni

---
Jeni Tennison
http://www.jenitennison.com/


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
Thread:
Adam Van Den Hoven
Jeni Tennison

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