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Re: Re: Finding the maximum depth from a node
- From: Greg Faron <gfaron at integretechpub dot com>
- To: xsl-list at lists dot mulberrytech dot com
- Date: Fri, 10 May 2002 14:42:04 -0600
- Subject: Re: [xsl] Re: Finding the maximum depth from a node
- Reply-to: xsl-list at lists dot mulberrytech dot com
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.
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