Question

我遇到了上述错误，并尝试打印出对象以查看我如何访问其中的数据，但它只回显DOMNodeList Object（）

function dom() {
$url = "http://website.com/demo/try.html";
$contents = wp_remote_fopen($url);

$dom = new DOMDocument();
@$dom->loadHTML($contents);
$xpath = new DOMXPath($dom);

$result = $xpath->evaluate('/html/body/table[0]');
print_r($result);
    }

我正在使用Wordpress，因此解释了wp_remote_fopen函数。我正试图从$ url

回显第一个表

Answer 1

是的，DOMXpath::query返回始终是DOMNodeList，这是一个奇怪的对象需要处理。您基本上必须迭代它，或者只使用item()来获取单个项目：

// There's actually something in the list
if($result->length > 0) {
  $node = $result->item(0);
  echo "{$node->nodeName} - {$node->nodeValue}";
} 
else {
  // empty result set
}

或者你可以遍历这些值：

foreach($result as $node) {
  echo "{$node->nodeName} - {$node->nodeValue}";
  // or you can just print out the the XML:
  // $dom->saveXML($node);
}

Answer 2

Xpath以1而不是0开始索引：/html/body/table[1]

现在，这取决于您是要保存匹配节点的HTML还是要保存节点的文本内容。

$html = <<<'HTML'
<html>
  <body>
    <p>Hello World</p>
  </body>
</html>
HTML;

$dom = new DOMDocument();
$dom->loadHTML($html);
$xpath = new DOMXPath($dom);

// iterate all matched nodes and save them as HTML to a buffer
$result = '';
foreach ($xpath->evaluate('/html/body/p[1]') as $p) {
  $result .= $dom->saveHtml($p);
}
var_dump($result);

// cast the first matched node to a string
var_dump(
  $xpath->evaluate('string(/html/body/p[1])')
);

演示：https://eval.in/155592

string(18) "<p>Hello World</p>"
string(11) "Hello World"

DOMNodeList类的对象无法转换为字符串

2 个答案: