Php 从XPath中删除特定表

我有这段代码，并用DOMXPath删除了一个给定的td

$html = file_get_contents('WebProxy.html');
$xml = new DOMDocument();
$xml->validateOnParse = true;
@$xml->loadHTML($html);

$xpath = new DOMXPath($xml);
$table =$xpath->query("//*[@id='proxylisttable']")->item(0);

// for printing the whole html table just type: print $xml->saveXML($table); 

$rows = $table->getElementsByTagName("tr");

foreach ($rows as $row) {
  $cells = $row -> getElementsByTagName('td');
  foreach ($cells as $cell) {
        echo $cell->nodeValue. . '<br>';
  }
}

---

我该怎么做呢，谢谢您的帮助

也许您可以通过排除不需要的单元格来调整xpath查询

例如：

<?php
$html = file_get_contents('WebProxy.html');
$xml = new DOMDocument();
$xml->validateOnParse = true;
@$xml->loadHTML($html);

$xpath = new DOMXPath($xml);
$cells =$xpath->query("//*[@id='proxylisttable']/tr/td[position() != 3 and position() != 6 and position() != 7 and position() != 8]");

foreach ($cells as $cell) {
    echo $cell->nodeValue . "<br>";
}

validateOnParse=true；
@$xml->loadHTML（$html）；
$xpath=newdomxpath（$xml）；
$cells=$xpath->query（“/*[@id='proxylistable']/tr/td[position（）！=3和position（）！=6和position（）！=7和position（）！=8]”；
foreach（$cells作为$cell）{
echo$cell->nodeValue。“
”；
}

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<?php
$html = file_get_contents('WebProxy.html');
$xml = new DOMDocument();
$xml->validateOnParse = true;
@$xml->loadHTML($html);

$xpath = new DOMXPath($xml);
$cells =$xpath->query("//*[@id='proxylisttable']/tr/td[position() != 3 and position() != 6 and position() != 7 and position() != 8]");

foreach ($cells as $cell) {
    echo $cell->nodeValue . "<br>";
}