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