在php中按属性选择xml节点

问题描述：

如何在不遍历每个子对象和每个属性/值的情况下，通过了解属性值来查找节点值?

How do I find the node value by knowing the attribute value without traversing through every child and every attribute/value ?

$dom = new DOMDocument;
$dom->load('test.xml');

$rows = $dom->getElementsByTagName('row');

foreach ($rows as $row) {

$header = VALUE OF <field name="header">
$text = VALUE OF <field name="text">

}

XML:

<resultset>
  <row>
    <field name="item">2424</field>
    <field name="header">blah blah 1</field>
    <field name="text" xsi:nil="true" />
    ...
    </row>

  <row>
    <field name="item">5321</field>
    <field name="header">blah blah 2</field>
    <field name="text">some text</field>
    ...
  </row>
</resultset>

答

最简单的方法是使用 DOMXPath::query ^docs

The simplest thing to do is use DOMXPath::query^docs

以下代码查找<row>节点内具有名称属性等于"header"的所有<field>节点:

The following code finds all the <field> nodes within <row> nodes that have a name attribute equal to "header":

$dom = new DOMDocument;
$dom->loadXML($str); // where $str is a string containing your sample xml
$xpath = new DOMXPath($dom);
$query = "//row/field[@name='header']";

$elements = $xpath->query($query);

foreach ($elements as $field) {
  echo $field->nodeValue, PHP_EOL;
}

使用您提供的示例xml，以上输出:

Using the sample xml you provide, the above outputs:

blah blah 1
blah blah 2

在php中按属性选择xml节点

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