想获取一个网站的网站内容，但总是获取不到该怎么办

<code class="php"><?php function request($url, $ck = '', $referer = '', $data = array()){
    $headers = array(
        'User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/45.0.2454.101 Safari/537.36'
    );

    $method = 'GET';

    if(count($data) > 0){
        $method = 'POST';
        $headers[] = 'X-Requested-With: XMLHttpRequest';
        $headers[] = 'Content-Type: application/x-www-form-urlencoded; charset=UTF-8';
    }

    //如果有传递 Cookie
    if($ck != ''){
        $headers[] = 'Cookie: ' . $ck;
    }

    //如果有传递 Referer
    if($referer != ''){
        $headers[] = 'Referer: ' . $referer;
    }

    $opts = array(
        'http' => array(
            'method'=> $method,
            'header'=> implode("\r\n", $headers)
        )
    );

    if(count($data) > 0){
        $opts['http']['content'] = http_build_query ($data);
    }

    $context = stream_context_create($opts);

    $html = file_get_contents($url, false, $context);

    return array(
        $html,//本次请求得到的HTML
        $http_response_header//本次请求服务器返回的响应头
    );
}

//先请求一次, 从响应头中获取 Cookie
$data = request('http://www.reg007.com/');

$headers = implode("", $data[1]);

preg_match_all('/Set-Cookie: (.+?;)/', $headers, $session);

if(count($session) !== 2){
    die('获取Cookie失败!');
}

$ck = implode(' ', $session[1]);//得到Cookie

$data = request('http://www.reg007.com/search?q=981267080-at-qq.com', $ck, 'http://www.reg007.com/');

$html = $data[0];//取出来 HTML

preg_match('/var h="(.+?)"/', $html, $h);

if(count($h) !== 2){
    die('获取Ajax请求Token失败!');
}

$h = $h[1];

$ck .= ' q=' . urlencode('981267080@qq.com');

//这个查询比较耗时, 会有点慢
$data = request(
    'http://www.reg007.com/search/ajax',
    $ck,
    'http://www.reg007.com/',
    array(
        'q'=>'981267080@qq.com',
        'h'=>$h,
        'i'=>0,
        't'=>0
    )
);

$result = json_decode($data[0]);

var_dump($result);</code>