我有一个如下所示的 xml 文档:
<dict>
<word>
<sense>
<definition> This is the text of the definition.
<example>
<quote>This is the text of an example.</quote>
</example>
<source>
<place>This is the name of the place recorded</place>
</source>.
</definition>
</sense>
</word>
</dict>
我需要使用 xQuery 来转换它,使得 <example>
它的 child 成为 <definition>
的 sibling ,而<source>
它的 child 应该成为 <example>
的 child 。换句话说,我需要这个作为输出:
<word>
<sense>
<definition> This is the text of the definition. </definition>
<example>
<quote>This is the text of an example.</quote>
<source>
<place>This is the name of the place recorded.</place>
</source>
</example>
</sense>
</word>
正如您所看到的,原始 <source>
后面的句号也存在问题。需要成为 <place>
关闭之前的最后一个字符串的元素.
我已经创建了一个 xQuery 文件并弄清楚了如何从层次结构中删除元素,但我在递归处理节点和在同一函数中添加新元素时遇到了麻烦。
xquery version "3.0";
declare namespace saxon="http://saxon.sf.net/";
declare option saxon:output "indent=yes";
declare option saxon:output "saxon:indent-spaces=3";
declare function local:test($node as item()*) as item()* {
typeswitch($node)
case text() return normalize-space($node)
case element(word) return <word>{local:recurse($node)}</word>
case element(dict) return <dict>{local:recurse($node)}</dict>
case element(sense) return <sense>{local:recurse($node)}</sense>
case element(definition) return local:definition($node)
case element(example) return local:example($node)
case element(source) return local:source($node)
case element(place) return <place>{local:recurse($node)}</place>
default return local:recurse($node)
};
declare function local:definition($nodes as item()*) as item()*{
(: here I need to process children of definition - except <source> and its
children will become children of <example>; and <example> should be returned
as a next sibling of definition. THIS IS THE PART THAT I DON'T KNOW HOW TO DO :)
<definition>
{
for $node in $nodes/node()
return
local:test($node)
}
</definition>
};
declare function local:example($node as item()*) as item()* {
(: here i am removing <example> because I don't want it to be a child
of <definition> any more. THIS BIT WORKS AS IT SHOULD :)
if ($node/parent::definition) then ()
else <example>{$node/@*}{local:recurse($node)}</example>
};
declare function local:source($node as item()*) as item()* {
(: here i am removing <source> because I don't want it to be a child
of <definition> any more. :)
if ($node/parent::definition) then ()
else <example>{$node/@*}{local:recurse($node)}</example>
};
declare function local:recurse($nodes as item()*) as item()* {
for $node in $nodes/node()
return
local:test($node)
};
local:test(doc("file:test.xml"))
这应该不是一件非常困难的事情,但我在 xQuery 如何处理此类问题方面遇到了概念上的困难。我将非常感谢您的帮助。
XSLT 不是一个选项。
最佳答案
为了完整起见,这里提供了一个仅包含一个递归函数的递归 XQuery 1.0 解决方案。我同意 Jens 的观点,即给定的示例可以很容易地处理,无需递归,但如果实际示例更大,并且您没有 XQuery Update 可供使用,您可以尝试如下操作:
declare function local:recurse($node as item()*) as item()* {
typeswitch($node)
case text()
return normalize-space($node)
case element(definition)
return element {node-name($node)} {
$node/@*,
local:recurse($node/node() except $node/(example|source))
}
case element(sense)
return element {node-name($node)} {
$node/@*,
local:recurse($node/node()),
<example>{
$node/definition/example/@*,
$node/definition/example/node(),
$node/definition/source
}</example>
}
case element()
return element {node-name($node)} {
$node/@*,
local:recurse($node/node())
}
default return $node
};
let $xml :=
<dict>
<word>
<sense>
<definition> This is the text of the definition.
<example>
<quote>This is the text of an example.</quote>
</example>
<source>
<place>This is the name of the place recorded.</place>
</source>
</definition>
</sense>
</word>
</dict>
return local:recurse($xml)
呵呵!
关于xml - xQuery 更改节点层次结构(从一个节点中删除子节点并将其作为兄弟节点返回),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24235325/