Question

例如，给定XML：

<root>
    <item>
        <id>111</id>
        <description>aisle 12, shelf 3</description>
        <description>inside the box</description>
    </item>
</root>

我想结果：

<root>
    <item>
        <id>111</id>
        <description>aisle 12, shelf 3 inside the box</description>
    </item>
</root>

但是节点可以有任何名称，并且可以处于任何级别。我希望同样的查询可以使用不同的XML，只要重复标记：

<root>
    <item>
        <id>112</id>
        <attributes>
            <author>Joe Smith</author>
            <author>Arthur Clarke</author>
            <author>Jeremiah Wright</author>
        </attributes>
    </item>
</root>

输出：

<root>
    <item>
        <id>112</id>
        <attributes>
            <author>Joe Smith Arthur Clarke Jeremiah Wright</author>
        </attributes>
    </item>
</root>

BaseX可以实现吗？如果没有，我们可以在给定已知元素的情况下执行此操作（例如，仅针对/ root / item / attributes / author）吗？

Answer 1

确保仅在兄弟姐妹之后直接合并会使事情变得复杂。我在下面的代码中添加了一些评论。

let $xml := document{<root>
    <item>
        <id>112</id>
        <attributes>
            <author>Joe Smith</author>
            <author>Arthur Clarke</author>
            <author>Jeremiah Wright</author>
            <foo/>
            <author>Donald Duck</author>
        </attributes>
    </item>
</root>}
return
  (: Use an XQuery Update transformation :)
  copy $copy := $xml
  modify (
    (: Loop over all leaves (only containing text nodes. :)
    (: This might have to be adjusted if you want to merge arbitrary nodes. :)
    for $leaf in $copy//*[not(*)]
    (: Where the preceding node is not of the same name :)
    (: (as it will be merged anyway) :)
    where not($leaf/preceding-sibling::*[1 and name(.) eq name($leaf)])
    (: Now find following siblings... :)
    let $siblings := $leaf/following-sibling::*[
      (: ... of the same name ... :)
      name(.) eq name($leaf) and
      (: ... and that do not have a node with another name in-between :)
      not(preceding-sibling::*[name(.) != name($leaf) and $leaf << .])
    ]
    return (
      (: Merge text contents into $leaf :)
      replace value of node $leaf with string-join(($leaf, $siblings), ' '),
      (: And delete all others :)
      delete nodes $siblings
    )
  )
  return $copy

无论文档中的位置如何，Basex XQL都会使用相同的标记名称合并内容

1 个答案: