[xquery-talk] If Statements within an If Statement?

Wei, Alice J. ajwei at indiana.edu
Wed Apr 2 15:32:46 PST 2008

Hi, David:

  Thanks, this did do the trick.

Alice Wei
MIS 2008
School of Library and Information Science
Indiana University Bloomington
ajwei at indiana.edu
From: David Carlisle [davidc at nag.co.uk]
Sent: Wednesday, April 02, 2008 1:47 PM
To: Wei, Alice J.
Cc: talk at x-query.com
Subject: Re: [xquery-talk] If Statements within an If Statement?

You are passing in a sequence of ad elements,

so $seq[.=$d][1] is an ad element



is empty, as your ad elements don't have any ancestor at all, definitely
not an ancestor named ad.

and all the variables you declare in the let clause are empty.

perhaps you meant

let $head := $seq[.=$d][1]/descendant::head[1]

but that's just a guess.

If you do that then
let $para := $seq[.=$d][1]/descendant::p

will be a sequence of p elements so you don't want


which will put p in p, just





will not always return just one elemet, which appears to be what you
intent, it will return every p that is the first child of its pareent.


on the other hand just returns the first p in foo, which is why I
changed your examples aboev, in addition to removing the spurious
ancestor axis.

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