排除路徑鏈接的網址？

在下面的函數中，我想指定一個從結果中排除的域列表。有什麼選擇？數組集合要排除？排除路徑鏈接的網址？

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); 
    }

來源

2011-08-14 Scott B

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); 
    }

來源

2011-08-14 15:49:55

+1看起來不錯Jason。謝謝！ –

@Scott B很高興它幫助你，如果你覺得它有用，請接受它。 –

我收到一個錯誤>「第二個參數'myFilter'應該是一個有效的回調」 –

既然你標記這個XPath，這裏是如何與XPath contain function:

$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; 
}

做這將輸出：

http://baz.example.com

見Xpath 1.0. specification for other possible string functions測試節點集。

來源

2011-08-14 16:41:13 Gordon

排除路徑鏈接的網址？

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