XML 将值从父级更改为子级

Question

我有这个 XML：<destinos> <destino>   <location>Spain</location>     <programas>       <item></item>       <item></item>     </programas> </destino> <destino>   <location>France</location>     <programas>       <item></item>       <item></item>     </programas> </destino></destinos>我需要在每个“项目”中包含或复制“位置”的值，但我无法这样做。<destinos> <destino>   <location>Spain</location>     <programas>       <item>        <location>Spain</location>       </item>       <item>        <location>Spain</location>       </item>     </programas> </destino> <destino>   <location>France</location>     <programas>       <item>        <location>France</location>       </item>       <item>        <location>France</location>       </item>     </programas> </destino></destinos>我对PHP一无所知，我一直在阅读，但找不到解决方案。如果有人可以帮助我并解释我将非常感激。我的代码：$url = file_get_contents("archive.xml");$xml = simplexml_load_string($url); $changes = $xml->xpath("//*[starts-with(local-name(), 'item')]");foreach ($changes as $change)     $change[0] = $xml->destinos->destino->location;header('Content-Type: application/xml');echo $xml->asXML();

Answer 1

使用 DOM，您可以将位置节点的克隆附加到相应的项目元素。

$document = new DOMDocument();

$document->load($url);

$xpath = new DOMXpath($document);

// iterate the location child of the destino elements

foreach($xpath->evaluate('//destino/location') as $location) {

    // iterate the item nodes inside the same parent node 

    foreach ($xpath->evaluate('parent::*/programas/item', $location) as $item) {

        // append a copy of the location to the item

        $item->appendChild($location->cloneNode(TRUE));

    }

}

echo $document->saveXML();

Answer 2

使用 SimpleXML，您可以只使用对象表示法来访问文档的各种元素，这样就不再需要 XPath 并且还可以使代码更具可读性...

$url = file_get_contents("archive.xml");

$xml = simplexml_load_string($url);

foreach ($xml->destino as $destino) {

    // Process each item

    foreach ( $destino->programas->item as $item )  {

        // Set the location from the destino location value

        $item->location = (string)$destino->location;

    }

}

header('Content-Type: application/xml');

echo $xml->asXML();

需要注意的一点是，在使用 SimpleXML 时，根节点（<destinos>在本例中）是$xml对象。这就是$xml->destino访问<destino>元素的原因。

Answer 3

一种选择是将xpath与addChild与位置值一起使用：

$url = file_get_contents("archive.xml");

$xml = simplexml_load_string($url);

$changes = $xml->xpath("/destinos/destino");

foreach ($changes as $change) {

    $text = (string)$change->location;

    foreach ($change->xpath("programas/item") as $i) {

        $i->addChild("location", $text);

    }

}

header('Content-Type: application/xml');

echo $xml->asXML();

输出

<destinos>

    <destino>

        <location>Spain</location>

        <programas>

            <item><location>Spain</location></item>

            <item><location>Spain</location></item>

        </programas>

    </destino>

    <destino>

        <location>France</location>

        <programas>

            <item><location>France</location></item>

            <item><location>France</location></item>

        </programas>

    </destino>

</destinos>