xquery: iteration simulation

[Michalmas]

Hello guys,

thanks a lot!

Both solutions seems to be ok (i only had to fix some typos in Michael's
solution, like missing '"' or too many ')' :P). The main purpose is, as
Micheal guessed, histogram of some nodes.

Best,
M.



On Mon, Mar 9, 2009 at 12:29 AM, Michael Kay <http://x-query.com/mailman/listinfo/talk> wrote:

>  Is "someStr" the same element name throughout? If so, it looks like this:
>
> declare function local:depth($e as element(*)) as xs:integer {
>   count($e/ancestor::*[node-name(.)=node-name($e)]
> };
>
> declare function local:histogram($root as element(mainNode), $name as
> xs:QName) as element(count)*
>   for $d in distinct-values(mainNode//*[node-name()=$s]/local:depth(.))
>   return <count element="{$name}" depth="{$d}
> count="{count(mainNode//*[node-name()=$name][local:depth(.)=$d]))
> }
>
> for $n in (xs:QName("", "someStr"), xs:QName("", "someOtherStr"), .....)
> return local:histogram(mainNode, $n)
>
> Michael Kay
> http://www.saxonica.com/
>
>  ------------------------------
> *From:* http://x-query.com/mailman/listinfo/talk [mailto:http://x-query.com/mailman/listinfo/talk] *On
> Behalf Of *Michalmas
> *Sent:* 08 March 2009 20:29
> *To:* http://x-query.com/mailman/listinfo/talk
> *Subject:*  xquery: iteration simulation
>
> Hello guys,
>
> Can you help me with the following problem:
>
> I have some XML.
>
> Now, i need to query it. But what i need to do, is to make queries of the
> form:
>
> count(mainNode//someStr)
> count(mainNode//someStr/someStr)
> (...)
>
> count(mainNode//someStr/someStr/someStr/someStr/someStr/someStr/someStr/someStr)
>
> The deep level of the last query may be even around 50 (so, 50 queries).
> Such set of queries will be repeated for about 100 keywords (like someStr).
> In total, that gives about 500 queries to create.
>
> Is is posible to write iteration in XQuery for that. Or pass it as a
> function?
>
> Thanks,
> M.
>
>
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