This is the mail archive of the xsl-list@mulberrytech.com mailing list .


Index Nav: [Date Index] [Subject Index] [Author Index] [Thread Index]
Message Nav: [Date Prev] [Date Next] [Thread Prev] [Thread Next]
Other format: [Raw text]

Re: Re: Finding the maximum depth from a node


At 02:11 PM 5/10/2002, you wrote:
>Here's a very simple solution -- use FXSL and its maximum()
>function/template. Pass to it the list of "folder" nodes and a
>reference to your comparison function, which will return 1 if the
>"name" attribute of the first argument is bigger than the name
>attribute of the second argument.
>
>       <xsl:if test="$arg1/@name > $arg2/@name">1</xsl:if>

   This only compares the name attribute of the different nodes and returns 
that which has the highest value.  The original question was to return the 
child node that had the largest hierarchy length.  The OP might have 
confused the issue by making them the same node.

   I was thinking something like the following, but it depends on knowing 
the length.
   <xsl:value-of select="*//*[count(ancestor::*)>3]/@name"/>

   Too bad there's not an extension function that works like the following 
invalid XSLT code:
   <xsl:value-of select="*//*[max(count(ancestor::*))]/@name"/>

   If I think of anything, I'll post it.


Greg Faron
Integre Technical Publishing Co.



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


Index Nav: [Date Index] [Subject Index] [Author Index] [Thread Index]
Message Nav: [Date Prev] [Date Next] [Thread Prev] [Thread Next]