Distinct Result via xQuery - distinct

Question

I'm trying to get reviewers who review one or more books published after 2010.
for $r in doc("review.xml")//Reviews//Review,
$b in doc("book.xml")//Books//Book
where $b/Title = $r/BookTitle
and $b/Year > 2010
return {$r/Reviewer}
The following are both XML files.
review.xml:
<Reviews>
<Review>
<ReviewID>R1</ReviewID>
<BookTitle>B1</BookTitle>
<Reviewer>AAA</Reviewer>
</Review>
<Review>
<ReviewID>R2</ReviewID>
<BookTitle>B1</BookTitle>
<Reviewer>BBB</Reviewer>
</Review>
<Review>
<ReviewID>R3</ReviewID>
<BookTitle>B2</BookTitle>
<Reviewer>AAA</Reviewer>
</Review>
<Review>
<ReviewID>R4</ReviewID>
<BookTitle>B3</BookTitle>
<Reviewer>AAA</Reviewer>
</Review>
<Reviews>
book.xml:
<Books>
<Book>
<Title>B1</Title>
<Year>2005</Year>
</Book>
<Book>
<Title>B2</Title>
<Year>2011</Year>
</Book>
<Book>
<Title>B3</Title>
<Year>2012</Year>
</Book>
</Books>
I'll get two AAA by my xQuery code. I was wondering if I can get the distinct result, which means only one AAA. I've tried distinct-value() but don't know how to use it probably. Thanks for your reply!
----My Updated Solution with XML format for xQuery 1.0----
<root>
{
for $x in distinct-values
(
for $r in doc("review.xml")//Reviews//Review,
$b in doc("book.xml")//Books//Book
where $b/Title = $r/BookTitle
and $b/Year > 2010
return {$r/Reviewer}
)
return <reviewer>{$x}</reviewer>
}
</root>

Answer 1

To preserve nodes, you can use the "group by" clause and select the first item of a group sequence:
for $r in doc("review.xml")//Review,
$b in doc("book.xml")//Book
let $n := $r/Reviewer
where $b/Title = $r/BookTitle
and $b/Year > 2010
group by $n
return $r[1]/Reviewer

Answer 2

The following query will give you all distint reviewer names (note that the values are atomized, which means the element nodes are removed):
distinct-values(
for $r in doc("review.xml")//Reviews//Review,
$b in doc("book.xml")//Books//Book
where $b/Title = $r/BookTitle
and $b/Year > 2010
return $r/Reviewer
)