DOMNode->replaceChild()
tfg_allardyce at gmail dot com
I have had exactly this problem.
To rectify I've had to do this:
<?php
$old_element = $doc->getElementsByTagName('Element1')->item(0);
$new_element = $doc->createElement('NewElement1');
$old_element_childNodes = $old_element->childNodes;
$length = $old_element_childNodes->length;
for($i = 0; $i < $length; $i++)
{
$oldChildren_array[] = $old_element_childNodes->item($i);
}
foreach($oldChildren_array as $old_c)
{
$new_element->appendChild($old_c);
}
?>
Rather than this:
(which I will bug report)
<?php
$old_element = $doc->getElementsByTagName('Element1')->item(0):
$new_element = $doc->createElement('NewElement1');
foreach($old_element->childNode as $node)
{
$new_element->appendChild($node);
}
?>
Using the latter, randomally removes the children!
DOMNodelist->item()
(No version information available, might be only in CVS)
DOMNodelist->item() — Retrieves a node specified by index
Popis
class DOMNodeList {DOMNode item ( int $index )
}
Retrieves a node specified by index within the
DOMNodeList object.
Tip
If you need to know the number of nodes in the collection, use
the length property of the
DOMNodeList object.
Seznam parametrů
- index
Index of the node into the collection.
Návratové hodnoty
The node at the indexth position in the
DOMNodeList, or NULL if that is not a valid
index.
Příklady
Příklad 413. Traversing all the entries of the table
<?php
$doc = new DOMDocument;
$doc->load('book.xml');
$items = $doc->getElementsByTagName('entry');
for ($i = 0; $i < $items->length; $i++) {
echo $items->item($i)->nodeValue . "\n";
}
?>
Alternatively, you can use foreach, which is a much more convenient way:
<?php
foreach ($items as $item) {
echo $item->nodeValue . "\n";
}
?>
Výše uvedený příklad vypíše:
Title
Author
Language
ISBN
The Grapes of Wrath
John Steinbeck
en
0140186409
The Pearl
John Steinbeck
en
014017737X
Samarcande
Amine Maalouf
fr
2253051209
add a note
User Contributed NotesDOMNodelist->item()
james dot dunmore at gmai dot com
27-Jul-2007 08:57
27-Jul-2007 08:57
Geoffrey Thubron
26-May-2007 02:47
26-May-2007 02:47
@ tfg_allardyce at gmail dot com
You could loop through the list backwards, that way, you are only ever taking off the last item from the list, and hence wont have disrupted the order.
oliver dot christen at camptocamp dot com
13-Feb-2007 07:27
13-Feb-2007 07:27
NodeList are something annoying because you can't output the content with a simple print_r, so I did a little function that add all the node to a new empty DOMDocument and output it as a string.
Have fun.
<?php
public function domNodeList_to_string($DomNodeList) {
$output = '';
$doc = new DOMDocument;
while ( $node = $DomNodeList->item($i) ) {
// import node
$domNode = $doc->importNode($node, true);
// append node
$doc->appendChild($domNode);
$i++;
}
$output = $doc->saveXML();
$output = print_r($output, 1);
// I added this because xml output and ajax do not like each others
$output = htmlspecialchars($output);
return $output;
}
?>
tfg_allardyce at gmail dot com
10-Jan-2007 03:57
10-Jan-2007 03:57
Be careful when looping through a DOMNodeList and moving its nodes around, sometimes this will take that node off the DOMNodeList and sometimes it wont!
<?php
// let $nodes be node list and $parent be some other node
foreach($nodes as $node) {
$parent->appendChild($node);
}
?>
In some cases the $node will be taken off the list and the next iteration of the loop will be corrupted, skipping every other node in the list! In other cases the node will remain in the list and everything will be fine.
Generally if you've created the node list using a getElementsByTagName call or an XPath query then the nodes will stay on the list. If the node list comes from another nodes' childNodes property those child nodes will be shifted off the list whenever you call appendChild.