Hello Giuseppe,

Would this do what you want:

let $reference := (1,3,2,3)

for $a  in (1,2,3,5,3)
  where $a = $reference
  return
    <c>{$a}</c>

Depending on how your $reference looks like, e.g. if it has many duplicates, you can compact it by doing
let $reference := distinct-values((1,3,2,3))
either way the returned sequence will be identical.

Best regards,
Kristian Kankainen

Ühel kenal päeval, N, 26.11.2020 kell 02:28, kirjutas Giuseppe G. A. Celano:
Unfortunately this does not work because, if the second sequence has only one 3 (and the first has two 3), I will still get two 3, while I should get only one.


Or more FLOWR like
 for $a  in (1,2,3,5,3)
 where $a . = (1,3,2)
 group by $o := $a
 return
 <c>{$a}</c>






On 25. Nov 2020, at 09:11, Martin Honnen <
martin.honnen@gmx.de
> wrote:

Am 25.11.2020 um 08:39 schrieb Martin Honnen:
Am 25.11.2020 um 06:37 schrieb Giuseppe G. A. Celano:

I have to compare two sequences and find common items, irrespective of
their positions:

for $a  in (1,2,3,5,3)
for $u  in (1,3,2,3)
where $a = $u
group by $o := $a
return
<c>{$a}</c>

This returns
<c>1</c>
<c>2</c>
<c>3 3 3 3</c>

I would like 3 to be repeated only twice (i.e., each item in the first
sequence should pair only with one in the second sentence): is there an
XQuery "trick" for that in the FLOWR expression?

I think

for $a  in (1,2,3,5,3)[. = (1,3,2,3)]
group by $o := $a
return
<c>{$a}</c>

would do that

Or more FLOWR like
 for $a  in (1,2,3,5,3)
 where $a . = (1,3,2,3)
 group by $o := $a
 return
 <c>{$a}</c>