XQuery：如何返回每个元素中出现的节点

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>

我想检索每个出版商已经出版书籍的字段（在此示例中为“web”）。

伪代码：如果所有publishers = fieldX都返回fieldX

Answer 1

如果每个发布者只有一本书，如示例所示，您可以直接检查每个字段，如果所有图书都有该字段：

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

否则，您需要过滤图书并分别考虑每个发布商：

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 = .]