Question

我现在非常沮丧，因为我似乎不明白为什么将一个节点（包含所有孩子）从一个XML树复制到另一个XML树很困难。

我googled很多，似乎我必须使用SimpleXML读取树，然后将节点解析为DOM以某种方式导入/导出它们。是否应该有一个功能一次性这样做？

基本上我想从源所有processList-＆gt;处理条目复制到目标test-＆gt; processList-＆gt;

source:
<processList>
    <process ...>
        //some more child nodes
    </process>
</processList>

destination:
<test>
    <processList>
        <process ...>
            //some more child nodes
        </process>
    </processList>
</test>

有人可以提出建议吗？

Answer 1

DOM可以轻松完成此任务。 DOMDocument::importNode()允许从其他文档导入节点。可以像任何节点一样附加/插入导入的节点。

DOMXpath::evaluate()可用于从DOM中获取节点。

$sourceXml = <<<'XML'
<processList>
  <process from="source">
    //some more child nodes
  </process>
</processList>
XML;    
$targetXml = <<<'XML'
<test>
  <processList>
    <process from="target">
      //some more child nodes
    </process>
  </processList>
</test>
XML;

$source = new DOMDocument();
$source->loadXml($sourceXml);
$sourceXpath = new DOMXPath($source);

$target = new DOMDocument();
$target->preserveWhiteSpace = false;
$target->formatOutput = true;
$target->loadXml($targetXml);
$targetXpath = new DOMXPath($target);

$targetNode = $targetXpath->evaluate('//processList[1]')->item(0);    
foreach ($sourceXpath->evaluate('//process') as $process) {
  $targetNode->appendChild(
    // import the node into the target document
    $target->importNode($process, TRUE)
  );
}    
echo $target->saveXml();

输出：

<?xml version="1.0"?>
<test>
  <processList>
    <process from="target">
      //some more child nodes
    </process>
    <process from="source">
    //some more child nodes
  </process>
  </processList>
</test>

PHP XML节点应对

1 个答案: