Json 如何从这个php构建参数/查询？

我想从这段php代码中构建参数/查询

$url = 'https://link.com/search?url=';

///Try to fetch compressed content using the file_get_contents function
$opts = array(
'http'=>array(
    'method'=>"GET",
    'header'=>"Accept-language: en\r\n" .
              "Cookie: foo=bar\r\n" .  // check function.stream-context-create on php.net
              "User-Agent: Mozilla/5.0 (iPad; U; CPU OS 3_2 like Mac OS X; en-us) AppleWebKit/531.21.10 (KHTML, like Gecko) Version/4.0.4 Mobile/7B334b Safari/531.21.102011-10-16 20:23:10\r\n" // i.e. An iPad 
  )
);

$context = stream_context_create($opts);
$zipped_content = file_get_contents($url ,false,$context);

        
        $json = json_encode($zipped_content,true);
       
        $str = str_replace('\\', '/', $json);
        $str2 = str_replace('////', '//', $str);
        $str3 = str_replace('//', '/', $str2);
        $str4 = str_replace('https:/', 'https://', $str3);
        
        preg_match_all('!https://[a-z0-9\-\.\/]+\.(?:jpe?g|png|gif)!Ui' , $str4 , $result);

foreach ($result[0] as $url)
{
  
    echo $url . "<br />\n";
}

可以这样使用 localhost/test.php？什么东西=https://google.com

怎么做？ 另外，提前谢谢！ 对不起，我的英语不好

$url = 'https://link.com/search?url=".$stuff."';