Question

$some_link = 'http://www.example.com';
$abc = 'killer';
$bcd = 'awsome';
$cde = 'qwerty';

$dom = new DOMDocument;
$dom->preserveWhiteSpace = false;
@$dom->loadHTMLFile($some_link);

$html = getTags( $dom, $abc, $bcd, $cde );
echo $html;

function getTags( $dom, $abc, $bcd, $cde ){
    $html = '';
    $domxpath = new DOMXPath($dom);
    $newDom = new DOMDocument;
    $newDom->formatOutput = true;

    $defffff = $domxpath->query("//$abc" . '[@' . $bcd . "='$cde']");

    // since above returns DomNodeList Object
    //  converting to string(html)
    $i = 0;
    while( $myItem = $defffff->item($i++) ){
        $node = $newDom->importNode( $myItem, true );    // import node
        $newDom->appendChild($node);                    // append node
    }
    $html = $newDom->saveHTML();
    return $html;
}

?>

这是整个代码。它连续返回多个结果，现在我想要的只有结果no.1和no.5。我该怎么办？

我是DOM新手，尝试了几件事但没有成功。在此先感谢

Answer 1

更改此

$i = 0;
while( $myItem = $defffff->item($i++) ){
    $node = $newDom->importNode( $myItem, true );    // import node
    $newDom->appendChild($node);                    // append node
}

进入此，以append仅选择节点

$i = 0;
while( $myItem = $defffff->item($i++) ){
    if ($i==0 or $i==4){
        $node = $newDom->importNode( $myItem, true );    // import node
        $newDom->appendChild($node);                    // append node
    }
}

或者如果你知道你想要的索引，你可以这样做

$myIndexes = array (0,4);
foreach ($myIndexes as $i){
    $myItem = $defffff->item($i++);
    $node = $newDom->importNode( $myItem, true );    // import node
    $newDom->appendChild($node);                    // append node
}

DOM Xpath仅获得所需的结果

1 个答案: