如何在没有www的情况下显示网址名称？

Question

如何在没有www的情况下显示网址名称？

如果网址为

  

https://www.google.com/search/

如何制作这样的？

  

google.com

我的代码

interface ICommand
{
    public function execute(Game $game) : void;
}

class RollDiceCommand implements ICommand
{
    private $player;

    public function __construct(Player $player)
    {
        $this->player = $player;
    }

    public function execute(Game $game) : void
    {
        $dice = DiceFacade::roll(new NumberGenerator());

        // Currently a business logic goes here

        if ($dice->isDouble()) {
            $player->incrementDoubleCount();

            if ($player->getDoubleCount() === 3) {
                $command = new GoToJailCommand();
                $command->execute();

                return;
            }
        } else {
            // The next player turn
            $game->nextPlayer();
        }

        $command = MoveForwardCommand($this->player);
        $command->execute($dice->getValue());

        // ...
    }
}

Answer 1

$url = 'http://sub.example.com/path?googleguy=googley';
$test = parse_url($url);
$string = explode('.',$test["host"]);
$top_lev_dom = array("com","net","us","gov","io","xyz","org","int","edu");
$output;
$end = false;
$pos = 0;
$pos2;
for($i = 0; $i < count($string); $i++){
    for($x = 0; $x < count(top_lev_dom); $x++){
        if($string[$i]==$top_lev_dom[$x]){
            $pos = $i;
            $pos2 = $i - 1;
            $end = true;
            break;
        }
        if($end == true){break;}
    }
    if($end == true){break;}
}

if($pos==0){
    $output = "error";
}else{
    $output.=$string[$pos2].".".$string[$pos];
}

echo $output;

  

example.com