2012-08-23 144 views
0

我只能从某些网站列表中获得域名。除了.com,www和子域之外,我只需要购买商店名称。同样,我必须只采取一个商店名称,并插入分区只得到.com域名和子域名

例如。该列表是

shopping.indiatimes.com 
shopping.indiatimes.com 
shopping.indiatimes.com 
www.infibeam.com 
adexmart.com 
deals.sulekha.com  
deals.sulekha.com 
www.ebay.in 
deals.ebay.in 
shop.seventymm.com 
www.homeshop18.com 
www.buytheprice.com 
www.snapdeal.com 
www.aircel.com 
www.tradus.com 
www.infibeam.com 
www.futurebazaar.com 
www.tradus.com 
www.shoppingmantra.com 
camera.zoomin.com 
www.homeshop18.com 

所以在这里我只想域名一样,indiatimes,易趣,tradus,zoomin..etc。

+0

'$域= parse_url($行[ '附加链接']); $ host = $ domain ['host']; $ exp = explode('。',$ host); $ domainname =($ exp [1]); '但我得到许多相同的商店。我想只带一个商店和插图 –

回答

0
$domain = parse_url("www.domainnames.com"); 
    $host = $domain['host'];  
    $exp = explode('.' , $host);   
       if(isset($exp[2])){   
       $domainname = ($exp[1]); 
       }else {   
        $domainname = ($exp[0]); 
       } 
       $a []= $domainname; 

     $uniquenames = (array_unique($a));  
     print_r($uniquenames); 
1
foreach($domains as $domain) 
    echo implode('', array_slice(explode('.', $domain), -2, 1)); 
0

这可以通过做一个字符串爆炸然后取第二个数组元素来完成。

$array = explode(".", "www.snapdeal.com"); 

$count = count($array); 
$value = $array[$count - 2] 
0
$url_to_check = "http://www.shopping.indiatimes.com"; 
$url1 = preg_replace('#^https?://#', '', $url_to_check); 
$domain = implode('', array_slice(explode('.', $url1), -2, 1)); 
echo $domain; //this will echo indiatimes