xml - 如何将 xpath 传递到 xquery 函数声明中

Question

我使用 Apache Tomcat 的 Exist DB 作为 XML 数据库，并尝试通过传递以下 xpath(在 FLWOR 的“let”子句中定义)来构造序列:

$xpath := $root/second/third

进入本地定义的函数声明，如下所示:

declare function local:someFunction($uuid as xs:string?, $xpath as xs:anyAtomicType?)
{
  let $varOne := $xpath/fourth[@uuid = $uuid]/fifthRight
  let $varTwo := $xpath/fourth[@uuid = $uuid]/fifthLeft
  let $combined := ($varOne,$varTwo)
  return $combined
};

当然，当在现有的 xquery 沙箱中输入此内容时，我得到 Type: xs:anyAtomicType is not Defined。我应该用什么来代替它，或者我应该以不同的方式来做这件事？

预先感谢您的任何建议。

Answer 1

我无法重现该错误(xs:anyAtomicType 未定义)。但是，也许以下内容可以提供帮助？

如果 $xpath (最初是一个节点)作为原子类型参数传递(因此被原子化)，那么当您尝试在函数中导航时，它肯定会抛出类型错误 XPTY0019 ($xpath/fourth)。以下代码在您这边工作吗(作为 node()* 传递)？

declare function local:someFunction($uuid as xs:string?, $xpath as node()*)
{
  let $varOne := $xpath/fourth[@uuid = $uuid]/fifthRight
  let $varTwo := $xpath/fourth[@uuid = $uuid]/fifthLeft
  let $combined := ($varOne,$varTwo)
  return $combined
};

let $root :=
  <first>
    <second>
      <third>
        <fourth uuid="1">
          <fifthLeft>foo</fifthLeft>
          <fifthRight>bar</fifthRight>
        </fourth>
      </third>
    </second>
  </first>
let $xpath :=$root/second/third
return
local:someFunction("1", $xpath)

(编辑:忘记星号以允许任意数量的节点)