[xquery-talk] step expression
Jonathan Robie
jonathan.robie at datadirect.com
Mon Apr 17 14:24:08 PDT 2006
Tim Finney wrote:
> This kind of expression is allowed:
>
> doc($doc_name)/node_seq_1/node_seq_2
>
> I would like to be able to do this:
>
> let
> $step1 := some_fn($name_space1, $element_name_1)
> $step2 := some_fn($name_space2, $element_name_2)
> return
> doc($doc_name)/$step1/$step2
>
> Can this be done?
>
Hi Tim,
You do this using the name() function:
let $doc_name := "invoice.xml"
let $step1 := "invoice"
let $step2 := "customer"
return
doc($doc_name)/*[name() eq $step1]/*[name() eq $step2]
Hope this helps!
Jonathan
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