Better way to structure XQuery

[Adam Retter]

2009/7/8  <http://x-query.com/mailman/listinfo/talk>:
> I have the following XQuery:
>
> xquery version="1.0";
>
> let $mydoc :=doc('http://localhost:8080/exist/rest/db/bio.xml')
> let $profileType := 'Standard'
> let $profileType := if
> ($mydoc/metadata/idinfo/spdom/bounding/boundalt/altmin or
> $mydoc/metadata/idinfo/spdom/bounding/boundalt/altmax) then 'Biological'
> else $profileType
>
> return
>   <result>
>      {$profileType>
>   </result>

This could possibly be rewritten in a couple of different ways. These
30+ paths, is there some common element to them - are they all
children of another node? are there other children at the same level?
I am thinking that you may be able to express your query without
evaluating 30 paths!

> Can anyone let me know how to use two variables, one being the $mydoc
> and one being a string ($path :=
> '/metadata/idinfo/spdom/bounding/boundalt/altmin') and have it evaluate
> as XPath instead of as a string?

You need an evaluate function. Unfortunately there is no evaluation
function in the XQuery 1.0 specification. However, almost every XQuery
implementation I have seen provides its own proprietary evaluation
function.  What XQuery implementation are you using?



-- 
Adam Retter

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