Xquery从两个XML DB中获取不同的数据

Question

以下是2个XML DB的示例结构-

主题数据库

<Subjects>
 <Subject>Physics</Subject>
 <Subject>Biology</Subject>
 <Subject>Maths</Subject>
 <Subject>Algebra</Subject>
</Subjects>

分类数据库

<Classification>
 <Subject>Physics</Subject>
 <Subject>English</Subject>
 <Subject>History</Subject>
 <Subject>Algebra</Subject>
</Classification>

我需要将两个数据库的不同值作为单个输出，并且排序也要像这样-

 <Subject>Algebra</Subject>
 <Subject>Biology</Subject>
 <Subject>English</Subject>
 <Subject>History</Subject>
 <Subject>Maths</Subject>
 <Subject>Physics</Subject>

Answer 1

经过一番尝试和错误之后，我能够获得所需的输出-

let $items := ( doc("Subject")//Topic , doc("Classification")//Topic )
for $x in distinct-values($items )
   order by $x
   return <Subject>{$x}</Subject>

我很困惑在哪里放置distinct-values()以及如何一次打开两个文档。