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我正在尝试使用 XPath 打印复杂的 XML 节点值，我附上了一张图片以帮助查看我需要到达的路径（红色下划线）。 XML 图像 原始 XML 文件可以在这里找到

我正在尝试这样的事情：

<?php
$xml = simplexml_load_file('document.xml');

    echo "<strong>Using direct method...</strong><br />";
    $names = $xml->xpath('/w:document/w:body/w:tbl[0]/w:tr[1]/w:tc[0]/w:p/w:r/w:t');
    foreach($names as $name) {
        echo "Found $name<br />";
    }

?>

我用来替换这个节点的这个方法：

 $file = "document.xml";                                        
    $fp = fopen($file, "rb") or die("error");
    $str = fread($fp, filesize($file));
    $xml = new DOMDocument();                                
    $xml->formatOutput = true;
    $xml->preserveWhiteSpace = false;
    $xml->loadXML($str) or die("Error");

    $root   = $xml->documentElement;
    $fnode  = $root->childNodes->item(0);

    $ori    = $fnode->childNodes->item(1);                      
    $ori1    = $ori->childNodes->item(3);
    $ori2   = $ori1->childNodes->item(1);
    $ori3   = $ori2->childNodes->item(1);
    $ori4   = $ori3->childNodes->item(1);
    $ori5   = $ori4->childNodes->item(1);
        $wt     = $xml->createElement("w:t");
    $wtText = $xml->createTextNode("".$name." ".$item."");
    $wt->appendChild($wtText);
        $ori4->replaceChild($wt,$ori5);
        $xml->save("document.xml");
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1 回答 1

1 
<?php

// Load XML
$doc = new DOMDocument();
$doc->load("document.xml");

// Use xpath to grab the node in question. I copied your xpath
// query as-is, assuming it was capable of targetting exactly
// the node you are trying to replace. If it returns more than
// one node, then only the first will be replaced.
// If this isn't what you want, I suggest modifying your xpath
// query to match exactly the single node you want to replace.
$xpath = new DOMXPath($doc);
$oldElement = $xpath->query("/w:document/w:body/w:tbl[0]/w:tr[1]/w:tc[0]/w:p/w:r/w:t")->item(0);
$newElement = $doc->createElementNS("http://schemas.openxmlformats.org/wordprocessingml/2006/main", "w:t", $name . " " . $item);

// Replace old element with new element
$oldElement->parentNode->replaceChild($newElement, $oldElement);

?>
于 2012-05-17T14:43:14.517 回答