How to find unique elements with XQuery? - xquery

Question

How could I get the list of title books where country location is unique in the next XML?
<BooksLib>
<Book Title="Murder in NY" Year="1980">
<BookLocations>
<Location City="New York" Country="USA"/>
<Location City="Virginia" Country="USA"/>
</BookLocations>
</Book >
<Book Title="Dracula" Year="2000">
<BookLocations>
<Location City="Sydney" Country="Australia"/>
<Location City="Moab" Country="USA"/>
<Location City="Calvados" Country="France"/>
</BookLocations>
</Book>
<Book Title="Romance in calvados" Year="2012">
<BookLocations>
<Location City="Calvados" Country="France"/>
</BookLocations>
</Book >
</BooksLib>
For example, in this XML would be "Dracula", because Australia only appears once
Now I got this:
for $book in doc("books.xml")//Book
where count(distinct-values($Book/BookLocations/Location/#Country)) eq 1
return $Book/data(#Title)
But this gives me the titles where the county is the same.

Answer 1

I would do it slightly differently. First, because it feels more natural to me how one actually thinks and second, because it will be faster on a large set. So first I would try to identify which countries are just present once. And then based on this filter out the results. The following should work:
let $single-loc :=
for $loc in doc("books.xml")//Location/#Country/string()
where count(//Location/#Country[. = $loc]) = 1
return $loc
for $book in doc("books.xml")//Book
where $book/BookLocations/Location/#Country = $single-loc
return $book/data(#Title)
Please note, that your input XML was not well-formed (I edited your post). Also, your XQuery is wrong as $book and $Book are two different variables.