XQuery：如何返回每個元素中出現的節點

Question

<bookstore>

<book category="Programming">
  <title lang="en">Coding</title>
  <publisher>ErBooks</publisher>
  <field>web</field>
  <field>programming</field>
  <field>C++</field>
</book>

<book category="XML">
  <title lang="en">Hey XML</title>
  <publisher>BookyBooks</publisher>
  <field>web</field>
  <field>xml</field>
  <field>database</field>
</book>

<book category="WEB">
  <title lang="en">XQuery Kick Start</title>
  <publisher>Penguin</publisher>
  <field>web</field>
  <field>design</field>
  <field>database</field>
</book>

我想檢索每個出版商已經出版書籍的字段（在此示例中為“web”）。

偽代碼：如果所有publishers = fieldX都返回fieldX

Answer 1

如果每個發布者只有一本書，如示例所示，您可以直接檢查每個字段，如果所有圖書都有該字段：

/book[1]/field[every $book in /book satisfies $book/field = .]

否則，您需要過濾圖書並分別考慮每個發布商：

let $bookstore := /bookstore  
let $publisher := distinct-values($bookstore/book/publisher)
let $fields := distinct-values($bookstore/book/field) 
return $fields[every $p in $publisher satisfies $bookstore/book[publisher = $p]/field = .]