如何通過xpath獲取每個節點的屬性?如何通過xpath獲取每個節點的屬性
例如,
是book.xml,
PHP,
<?php
$doc = new DOMDocument;
$doc->load('book.xml');
$xpath = new DOMXPath($doc);
# get and output "<entry>" elements
$x = $doc -> getElementsByTagName('record');
# Count the total feed with xpath.
$total = $x->length;
# the query is relative to the records node
$query = 'string(/records/@timestamp)';
for ($i=0; $i<$total; $i++)
{
$timestamp = $xpath->evaluate($query,$x->item($i));
echo $timestamp ."<br/>";
}
?>
結果(它僅循環的第一個節點),
1264777862
1264777862
1264777862
1264777862
但我想得到,
1264777862
1264777000
我已按照here的問題和答案進行了修改。
或者也許有更好的方法?
編輯:
XML,
<?xml version="1.0" encoding="UTF-8" ?>
<records>
<record timestamp="1264777862">A</record>
<record>B</record>
<record timestamp="1264777000">C</record>
<record>D</record>
</records>
與此,
for ($i=0; $i<$total; $i++)
{
$value = $x->item($i)->childNodes->item(0)->nodeValue;
$timestamp = $xpath->evaluate($query,$x->item($i));
echo $value.': '.$timestamp ."<br/>";
}
我得到這樣的結果,
A: 1264777862
B: 1264777862
C: 1264777862
D: 1264777862
但這是結果我之後,
A: 1264777862
B:
C: 1264777862
D:
編輯:
測試,
$nodes = $xpath->query('//records/record');
foreach($nodes as $node) {
$value = $node->nodeValue;
$timestamp = $node->getAttribute('timestamp');
echo $value .': '."<br/>";
}
結果,
A:
B:
C:
D:
你的XML對'records',一個在'record'的屬性。你想和誰一起工作? –
對不起,我的錯誤。請參閱我上面的編輯。謝謝。 – laukok