読みたいdivを取得するXPathがあります。1つのXPathでdivのクラスと名前を読み取る方法は?
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2 に答える
1
<?php
$xmlDoc = new DOMDocument();
$xmlDoc->load( '<<your file>>>' );
//$xmlDoc->loadHTML($sourceString); //-> if its a string you have
$xpath = new DOMXpath($doc);
$elements = $xpath->query("Your XPath");
//if you are sure there is only one div, for the Xpath, you can use a index 0 in the next statement, else uou have to itereate it in a loop
$node = $elements->item(0);
$attrib1 = $node->attributes->getNamedItem("<attribute_name1>");
$attrib2 = $node->attributes->getNamedItem("<attribute_name2>");
$attrib3 = $node->attributes->getNamedItem("<attribute_name3>");
....
?>
于 2012-06-05T13:47:09.507 に答える
0
このコードは、属性nameとclassを持つすべてのdivを選択します。選択したノードから情報を選択する方法については、以下を参照してください。
<?php
$xmlstring = '<root>' .
'<div name="value1" class="value2" myarg="value3"></div>' .
'<div name="value4"></div>' .
'</root>';
$xml = simplexml_load_string($xmlstring);
// Select all div's that have attributes name and class
$xpath = "//div[@name and @class]";
var_dump($result = $xml->xpath($xpath));
/* Outputs:
array(1) {
[0]=>
object(SimpleXMLElement)#3 (1) {
["@attributes"]=>
array(3) {
["name"]=>
string(6) "value1"
["class"]=>
string(6) "value2"
["myarg"]=>
string(6) "value3"
}
}
}
*/
// note that only one iteration is performed
// as the second div does not have an attribut
// called 'class'.
foreach($result as $element)
{
echo $element['name']; // value1
echo $element['class']; // value2
}
// or, if only one div is present in the document:
echo $result[0]['name']; // value1
echo $result[0]['class']; // value2
?>
于 2012-06-05T18:46:02.683 に答える