DOMNode::replaceChild

(PHP 5, PHP 7, PHP 8)

DOMNode::replaceChild — 子を置き換える

説明

public DOMNode::replaceChild(DOMNode $node, DOMNode $child): DOMNode|false

この関数は、子要素 child を新しいノードに置き換えます。もし node がすでに子要素であった場合は、それがふたたび追加されることはありません。置き換えに成功すると、置き換え前のノードが返されます。

パラメータ

node: 新しいノード。対象ドキュメントのメンバ、すなわち、DOMDocument->createXXX() メソッドのひとつで作成されたものか DOMDocument::importNode でドキュメントにインポートされたものである必要があります。
child: 古いノード。

戻り値

古いノード、あるいはエラーが発生した場合は false を返します。

エラー / 例外

DOM_NO_MODIFICATION_ALLOWED_ERR: このノードが読み込み専用であったり、挿入されるノードの以前の親が読み込み専用であった場合に発生します。
DOM_HIERARCHY_REQUEST_ERR: node で指定した型の子ノードを保持することが許可されていない場合、あるいは追加しようとしているノードが自分自身やその祖先であった場合に発生します。
DOM_WRONG_DOCUMENT_ERR: node が、このノードとは別のドキュメントで作成されたものである場合に発生します。
DOM_NOT_FOUND_ERR: child がこのノードの子でない場合に発生します。

参考

DOMChildNode::replaceWith() - ノードを新しいノードで置き換える
DOMNode::appendChild() - 子要素群の最後に新しい子要素を追加する
DOMNode::removeChild() - 子要素群から子要素を削除する

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User Contributed Notes 4 notes

down

jb at jbpros dot com ¶

19 years ago

If you are trying to replace more than one node at once, you have to be careful about iterating over the DOMNodeList.  If the old node has a different name from the new node, it will be removed from the list once it has been replaced.  Use a regressive loop:



<?php



$xml = new DOMDocument;

$xml->load('docfile.xml');



$elements = $xml->getElementsByTagNameNS('http://www.example.com/NS/', '*');

$i = $elements->length - 1;

while ($i > -1) {

    $element = $elements->item($i);

    $ignore = false;



    $newelement = $xml>createTextNode('Some new node!');

    $element->parentNode->replaceChild($newelement, $element);

    $i--;

}



?>



The loop counter ($i) will always be in the list's interval as removed elements indexes are above the counter.

down

aidan at php dot net ¶

19 years ago

Here is a simple example for replacing a node:

Let's define our XML like so:

<?php
$xml = <<<XML
<?xml version="1.0"?>
<root>
  <parent>
     <child>bar</child>
     <child>foo</child>
  </parent>
</root>
XML;
?>

If we wanted to replace the entire <parent> node, we could do something like this:

<?php
// Create a new document fragment to hold the new <parent> node
$parent = new DomDocument;
$parent_node = $parent ->createElement('parent');

// Add some children
$parent_node->appendChild($parent->createElement('child', 'somevalue'));
$parent_node->appendChild($parent->createElement('child', 'anothervalue'));

// Add the keywordset into the new document
// The $parent variable now holds the new node as a document fragment
$parent->appendChild($parent_node);
?>

Next, we need to locate the old node:

<?php
// Load the XML
$dom = new DomDocument;
$dom->loadXML($xml);

// Locate the old parent node
$xpath = new DOMXpath($dom);
$nodelist = $xpath->query('/root/parent');
$oldnode = $nodelist->item(0);
?>

We then import and replace the new node:

<?php
// Load the $parent document fragment into the current document
$newnode = $dom->importNode($parent->documentElement, true);

// Replace
$oldnode->parentNode->replaceChild($newnode, $oldnode);

// Display
echo $dom->saveXML();
?>

Our new node is successfully imported:

<?xml version="1.0"?>
<root>
<parent><child>somevalue</child><child>anothervalue</child></parent>
</root>

down

franp at free dot fr ¶

20 years ago

1) If your XPath query returns a NodeList including a unique item, or if you know for sure the order of the items returned, you can use the "item(n)" syntax instead of the "foreach" syntax.

This can greatly improve you code lisibility.

s the method name implies, replaceChild cannot replace a node itself but a child of a node.

The trick is to use replaceChild on the parent node of your Xpath query result.



<?xml version="1.0" ?>

<action>

 <actionGlobal>

  <actionGlobalFR>sometext</actionGlobalFR>

 <actionGlobal>

</action>



$frag = $doc->createElement("actionGlobalFR");   

$fragA = $doc->createTextNode("anothertext");

$frag->appendChild($fragA);

      

$xpResult = $xp->query("/action/actionGlobal/actionGlobalFR");   

$blipblip = $xpResult->item(0)->parentNode->replaceChild($ajout, $xpResult->item(0));



2) Also, as the method name implies, replaceChild cannot replace a node itself but a child of a node.

Still it is possible to replace a node pointed by XPath istead of its child. 

The trick is to use replaceChild on the parent node of your Xpath query result.



<?xml version="1.0" ?>

<action>

  <FR>French Text</FR>

</action>



<?php

$frag = $doc->createElement("EN");   

$fragA = $doc->createTextNode("English Text");

$frag->appendChild($fragA);

      

$xpResult = $xp->query("/action/FR");   

$blipblip = $xpResult->item(0)->parentNode->replaceChild($fragA, $xpResult->item(0));

?>



Et voil� !



This produces :

<?xml version="1.0" ?>

<action>

  <EN>English Text</EN>

</action>



....................................................



3) Also, be carefull, you CANNOT replace a node that doesn't exist.

While this may seems obvious, it is easy to forget.



Consider this :

<?xml version="1.0" ?>

<action>

  <EN></EN>

</action>



You cannot use replaceChild() to turn this into :

<?xml version="1.0" ?>

<action>

  <EN>Some text</EN>

</action>



The reason is that since the <EN></EN> element is empty, it has no child (this is clearer to understand if you consider that <EN></EN> can be written <EN />).

The fact that you intend to put some text inbetween <EN> and </EN> does not change the fact that it has no text yet, thus no child yet.

When dealing with DOM, do not take your dream for the reality. The DOM parser doen't care about your dreams. If an element is currently empty, it has no child, whatever you intend to fill in.



Thus, the solution to teh problem is to use appendChild intead of replaceChild :



<?php

$fragA = $doc->createTextNode("Some Text");

$xpResult = $xp->query("/action/EN");   

$blipblip = $xpResult->item(0)->appendChild($fragA);

?>



This produces the awaited :

<?xml version="1.0" ?>

<action>

  <EN>Some Text</EN>

</action>



....................................................



4) Note that the description of replaceChild in the doc is wrong. Arguments have been inverted.

The correct description is :

object DOMNode->replaceChild (object newnode, object oldnode)

down

chealer at gmail dot com ¶

6 years ago

As of version 5.6, PHP still behaves as reported by jb at jbpros dot com. To use replaceChild() in a loop, the more standard pattern used in the following example can be used:

for ($currentNode = 0; $currentNode < $link->childNodes->length; $currentNode++) {
    $child = $link->childNodes[$currentNode];

    // "Remove" links, since links can't contain links
    if ($child instanceof DOMElement && $child->tagName == 'a') {
        $replacement = $dom->createTextNode($child->textContent);
        $link->replaceChild($replacement, $child);
    }
}

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DOMNode::replaceChild

説明

パラメータ

戻り値

エラー / 例外

参考

Found A Problem?

User Contributed Notes 4 notes