Altova Mailing List Archives>Archive Index >xsl-list Archive Home >Recent entries >Thread Prev - [xsl] Finding deepest node [Thread Next] Re: [xsl] Finding deepest nodeTo: Date: 12/2/2005 9:00:00 PM Jon Gorman wrote: > Well......what do you mean by farthest away? I'm sure it's probably possible. When I say farthest away I mean the node that has the most nodes between itself and the context node in a straight line (not counting siblings). .//node[@a = 'avalue' and max(count(ancestor::*))]/@id I know that isn't a valid expression but I hope you know what I mean Example: <nodes> <node a="1" id="1"/> <node a="2" id="2"> <node a="1" id="3"> <node a="1" id="3.1"/> </node> <node a="1" id="4"/> </node> <node a="2" id="5"> <node a="1" id="6"> <node a="1" id="7"/> </node> </node> </nodes> Say that a node must have @a=1 to be selectable. In this example the nodes with @id=3.1 and @id=7 are furthest away from <nodes> and have a distance of 3. @id=3, @id=4, @id=6 have a distance of 2 and so on. /Marcus | ||||||
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