xml - 选择节点但忽略返回子节点的 XQuery

Question

我正在尝试创建一个 xquery 表达式，它将返回选定的节点但不会返回它们的子节点。这可能最好用一个例子来说明。我有以下 myNode 节点:

<myNode>
  <myElements id="1">
    <myElement key="one">aaa</myElement>
    <myElement key="two" >bbb</myElement>
    <myElement key="three">ccc</myElement>
  </myElements>
  <myElements id="2">
    <myElement key="one">ddd</myElement>
    <myElement key="two" >eee</myElement>
    <myElement key="three">fff</myElement>
  </myElements>
</myNode>

我有兴趣返回 节点，但仅此而已。我期望的返回如下所示:

<myNode>
  <myElements id="1" />
  <myElements id="2" />
</myNode>

或

<myNode>
  <myElements id="1"></myElements>
  <myElements id="2"></myElements>
</myNode>

我目前有一个类似于以下内容的 xpath 表达式(为此图进行了简化)，正如预期的那样，它将返回 myElement 子元素:

$results/myNode/MyElements

我是不是找错了树？这在 XPath/XQuery 中甚至可能吗？

Answer 1

试试这个递归算法..

    xquery version "1.0";

    declare function local:passthru($x as node()) as node()* {   for $z in $x/node() return local:recurseReplace($z) };

    declare function local:recurseReplace($x as node()) {
        typeswitch ($x) 
            (: Changes based on a condition :)
            case element(myElements) return <myElements id="{$x/@id}" />
            (: IGNORE ANY CHANGES :)
            case text() return $x
            case comment() return comment {"an altered comment"}
            case element() return element {fn:node-name($x)} {for $a in $x/attribute()
                                                              return $a, local:passthru($x)}
            default return ()           
};

    let $doc := 
<myNode>   
  <myElements id="1">
        <myElement key="one">aaa</myElement>
        <myElement key="two" >bbb</myElement>
        <myElement key="three">ccc</myElement>   
  </myElements>   
  <myElements id="2">
        <myElement key="one">ddd</myElement>
        <myElement key="two" >eee</myElement>
        <myElement key="three">fff</myElement>   
  </myElements> 
</myNode> 
return local:recurseReplace($doc)