[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: XQuery/XPath 3.1: Node List to Node Set ("distinct
Hmph. That is certainly much more efficient p but is not necessarily obvious (at least not to me). Thanks! E. _____________________________________________ Eliot Kimber Sr Staff Content Engineer O: 512 554 9368 M: 512 554 9368 servicenow.com<https://www.servicenow.com> LinkedIn<https://www.linkedin.com/company/servicenow> | Twitter<https://twitter.com/servicenow> | YouTube<https://www.youtube.com/user/servicenowinc> | Facebook<https://www.facebook.com/servicenow> From: Martin Honnen martin.honnen@xxxxxx <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> Date: Tuesday, December 28, 2021 at 1:15 PM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx <xsl-list@xxxxxxxxxxxxxxxxxxxxxx> Subject: Re: XQuery/XPath 3.1: Node List to Node Set ("distinct nodes") [External Email] On 28.12.2021 20:10, Eliot Kimber eliot.kimber@xxxxxxxxxxxxxx wrote: > I couldnbt find an answer in my google and markmail searching so I > thought Ibd ask here: > > Given an arbitrary list of nodes that may contain duplicates, what is > the most efficient way to reduce the node list to a set? > > The solution I came up with is a recursive function: > > (: > > Get the unique nodes from the supplied sequence > > @param nodes The sequence of nodes to evaluate > > @return A sequence of nodes such that each node in $nodes exists exactly > once. > > :) > > declare function dutils:distinctNodes($nodes as node()*) as node()* { > > dutils:_getDistinctNodes($nodes, ()) > > }; > > declare function dutils:_getDistinctNodes($nodes as node()*, $resultList > as node()*) as node()* { > > if (exists($nodes)) > > then > > let $node := head($nodes) > > return dutils:_getDistinctNodes(tail($nodes), ($resultList | $node)) > > else $resultList > > }; > > Which works but I feel like Ibm missing some obvious way to do this more > directly, but Ibm not seeing it. > > Am I missing a better solution? $nodes/.
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