本文介绍了从路径链接中排除网址?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
在下面的功能中,我想指定要从结果中排除的域列表。有哪些选择?
In the function below, I'd like to specify a list of domains to exclude from the results. What are some options? Array collection to exclude?
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);
}
推荐答案
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);
}
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