[xquery-talk] step expression

[Tim Finney]

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?

Best

Tim Finney
Programmer
Electronic Imprint
UVa Press