在php中按属性选择xml节点

Question

如何通过了解属性值而不遍历每个子项和每个属性/值来查找节点值？

$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>

Answer 1

最简单的方法是使用DOMXPath::query^docs

以下代码查找节点中名称属性等于"header"的所有<field>节点<row>:

$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,上面的输出:

blah blah 1
blah blah 2