[xquery-talk] Getting node position in a for statement

David Carlisle davidc at nag.co.uk
Mon Jun 16 15:17:01 PDT 2008


	let $pos := $rows/position()

$rows is (at each iteration) a single tr element so $rows/position()
will always be 1.

You want to use an "at" clause as in the for expression:

for $rows at $p in //table/tbody/tr
	let $cells := count($rows/th)
	let $title := $rows/ancestor::table/title...
<row position="{$p} cellcount="{$cells}" >{data($title)}</row>

or use an xpath rather than a for expression (so that position()
reflects the position in the sequence)

 //table/tbody/tr/<row position="{position()} cellcount="{count(th)}">

(This seems to be the same question, and same answer as last week


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