Question

是否可以只运行一个查询来替换多个查询？

 //host_info[hostname="localhost" and vulnerability_id="remote_execution"][1]
 //host_info[hostname="localhost" and vulnerability_id="remote_execution"][1]/@exclude
 //host_info[hostname="localhost" and vulnerability_id="remote_execution"][1]/../problem[1]
 //host_info[hostname="localhost" and vulnerability_id="remote_execution"][1]/../reference[1]
 //host_info[hostname="localhost" and vulnerability_id="remote_execution"][1]/../impact[1]
 //host_info[hostname="localhost" and vulnerability_id="remote_execution"][1]/../background[1]
 //host_info[hostname="localhost" and vulnerability_id="remote_execution"][1]/../resolution[1]

由于某种原因，这在PHP的xpath（）函数中不起作用。

//host_info[hostname="localhost" and vulnerability_id="remote_execution"]/(*, ../background, ../impact, ../problem)

我觉得这里必须有更好的方法。请注意，可能有多个<host_info>节点，这就是为什么我要指定主机名和脆弱性ID。但是父<background>节点中仅包含一个<resolution>，<vulnerability_id>节点。

<report>
<vulnerability>
   <host_info>
      <hostname></hostname>
      <vulnerability_id></vulnerability_id>
   </host_info>
   <host_info>
      <hostname></hostname>
      <vulnerability_id></vulnerability_id>
   </host_info>
   <background></background>
   <resolution></resolution>
</vulnerability>
<vulnerability>
   <host_info></host_info>
   <host_info></host_info>
   <host_info></host_info>
   <host_info></host_info>
   <background></background>
   <resolution></resolution>
</vulnerability>
</report>

Answer 1

节点列表只有一个维，因此将多个项目的详细信息序列化到一个列表中并没有真正的用处。

通常，您将使用一个Xpath表达式来标识您的列表节点，然后使用DOM方法和相关表达式来获取与这些节点相关的数据：

$document = new DOMDocument();
$document->loadXML($xml);
$xpath = new DOMxpath($document);

foreach ($xpath->evaluate('//host_info[hostname="localhost" and vulnerability_id="remote_execution"]') as $hostInfo) {
    var_dump(
      [
          $hostInfo->getAttribute('exclude'),
          $xpath->evaluate('string(parent::*/background)', $hostInfo),
          $xpath->evaluate('string(parent::*/resolution)', $hostInfo)
        ]
    );
}

对于单个项目，可以将所有详细信息节点提取到单个结果列表中。但是，这样的表达式很快就会变得复杂，然后您必须添加识别不同细节节点的逻辑。

$document = new DOMDocument();
$document->loadXML($xml);
$xpath = new DOMxpath($document);

$expression = <<<'XPATH'
//host_info[hostname="localhost" and vulnerability_id="remote_execution"][1]/@exclude|
(//host_info[hostname="localhost" and vulnerability_id="remote_execution"][1]/parent::*/background)[1]|
(//host_info[hostname="localhost" and vulnerability_id="remote_execution"][1]/parent::*/resolution)[1]
XPATH;

foreach ($xpath->evaluate($expression) as $detail) {
    var_dump(
       $detail->localName, $detail->textContent
    );
}

PHP XPath根据“子”值返回父节点

1 个答案: