RE: Finding corresponding node in structurally identic

> I need to quickly find the matching node to the current node
> which is in a parallel branch with identical structure to the
> branch I'm operating in. It seems like this should be really
> simple, but the only solution I can come up with is
> a recursive template to assemble the name of the path from
> the branch tag down, and then compare that against the result
> from the same template for each node in the second tree. The
> trees may be of arbitrary depth. Any other ideas?

Depends what you mean by "quickly", but you could try:

<xsl:variable name="branch1index"
select="count(/example/branch1/preceding::node())"/>
<xsl:variable name="thisindex" select="count(preceding::node())"/>

select="/example/branch2/descendant::node[$thisindex - $branch1index]"

Michael Kay
Software AG
home: Michael.H.Kay@xxxxxxxxxxxx
work: Michael.Kay@xxxxxxxxxxxxxx

>
> <example>
>   <branch1>
>      <somestuff>
>          <somedeeperstuff>
>              <someevendeeperstuff/> <--- Current node
>          </somedeeperstuff>
>      </somestuff>
>   </branch1>
>   <branch2>
>      <somestuff>
>          <somedeeperstuff>
>              <someevendeeperstuff/> <--- Corresponding node I
> want to find
>          </somedeeperstuff>
>      </somestuff>
>   </branch2>
>  </example>
>
> The reason we are doing this is to be able to cache a form
> description in the first branch.
> Then any default values are placed in the second branch which
> means that we don't need to modify the
> cached document fragment.
>
> TIA
>
> Joel
>
>  XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list
>


 XSL-List info and archive:  http://www.mulberrytech.com/xsl/xsl-list