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[XQuery Talk Mailing List Archive Home] [By Date] [By Thread] [By Subject] [By Author] [Recent Entries] [Reply To This Message] Construct the absolute xpath (with node position)Vallone, Philip Mr CTR USA AMC Philip.Vallone at us.army.milTue Jun 17 12:56:29 PDT 2008
Hi, is it possible to construct the absolute xpath (with node position) from an xquery expression? Take the following XML: <root> <item> <para>hello</para> <para>goodbye</para> </item> <item> <para>hello</para> <para>goodbye</para> <para>end</para> </item> <item> <para>hello</para> <para>middle</para> <para>goodbye</para> </item> </root> Take the following xquery: for $path in //item/para let $xpath := $path/string-join(ancestor-or-self::*/name(), '/') return <out>{$xpath}</out> Which returns: <out>root/item/para</out> <out>root/item/para</out> .... Is it possible to construct an xquery expression to output the xpath with node position: <out>/root/item[1]/para[1]</out> <out>/root/item[1]/para[2]</out> <out>/root/item[2]/para[1]</out> <out>/root/item[2]/para[2]</out> <out>/root/item[2]/para[3]</out> <out>/root/item[3]/para[1]</out> <out>/root/item[3]/para[2]</out> <out>/root/item[3]/para[3]</out> Thanks for the help. Phil
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