Question

在下面的函数中，我想指定要从结果中排除的域列表。有哪些选择？要排除的数组集合？

class KeywordSearch
{       
    const GOOGLE_SEARCH_XPATH = "//a[@class='l']";
    public $searchQuery;
    public $numResults ;
    public $sites;
    public $finalPlainText = '';
    public $finalWordList = array();
    public $finalKeywordList = array();

    function __construct($query,$numres=7){
        $this->searchQuery = $query;
        $this->numResults = $numres;
        $this->sites = array();
    }

    protected static $_excludeUrls  = array('wikipedia.com','amazon.com','youtube.com','zappos.com');//JSB NEW

    private function getResults($searchHtml){

        $results = array();
        $dom = new DOMDocument();
        $dom->preserveWhiteSpace = false;
        $dom->formatOutput = false;
        @$dom->loadHTML($searchHtml);
        $xpath = new DOMXpath($dom);
        $links = $xpath->query(self::GOOGLE_SEARCH_XPATH);

        foreach($links as $link)
        {
            $results[] = $link->getAttribute('href');           
        }

        $results = array_filter($results,'self::kwFilter');//JSB NEW
        return $results;
    }

    protected static function kwFilter($value)
    {
        return !in_array($value,self::$_excludeUrls);
    }

Answer 1

protected static $_banUrls  = array('foo.com','bar.com');

private function getResults($searchHtml){

        $results = array();

        $dom = new DOMDocument();

        $dom->preserveWhiteSpace = false;

        $dom->formatOutput = false;

        @$dom->loadHTML($searchHtml);

        $xpath = new DOMXpath($dom);

        $links = $xpath->query(self::GOOGLE_SEARCH_XPATH);


        foreach($links as $link)
        {
        //FILTER OUT SPECIFIC LINKS HERE
            $results[] = $link->getAttribute('href');

        }
        $results = array_filter($results,'self::myFilter');

        return $results;

    }

    protected static function myFilter($value)
    {
            return !in_array($value,self::$_banUrls);
    }

Answer 2

由于您标记了此XPath，以下是使用XPath contain function:

执行此操作的方法

$html = <<< HTML
<ul>
    <li><a href="http://foo.example.com">
    <li><a href="http://bar.example.com">
    <li><a href="http://baz.example.com">
</ul>
HTML;

$dom = new DOMDocument;
$dom->loadHtml($html);
$xp = new DOMXPath($dom);
$query = '//a/@href[
    not(contains(., "foo.example.com")) and
    not(contains(., "bar.example.com"))
]';
foreach ($xp->query($query) as $hrefAttr) {
    echo $hrefAttr->nodeValue;
}

这将输出：

http://baz.example.com

请参阅Xpath 1.0. specification for other possible string functions以测试节点集。

从路径链接中排除URL？

2 个答案: