php http请求问题

错误信息：{"errcode":41001,"errmsg":"access_token missing"}
//发送POST请求
$url = "https://api.weixin.qq.com/cgi-bin/qrcode/create?access_token=";
$access_token = "bz6LKNCiQN5fHDZNJwWbCiPXqRkrlkBUcBGwb3MlM-tmnXK6TGsHGbsETwcOXmezlIouHdD7Rv3g9aLicuF-gA";
$url = $url . urlencode($access_token);
echo "请求url：" . $url ."
";
//要请求的内容
$data['action_name'] = "QR_LIMIT_SCENE";
$scene['scene_id'] = 10;
$action_info['scene'] = $scene;
$data['action_info'] = $action_info;
$data = json_encode($data);
echo "请求参数：" . $data ."
";

//url
$url_info = parse_url($url);
var_dump($url_info);
echo "
";
if(!isset($url_info['port']))
{
$url_info['port'] = 80;
//模拟http请求头
$request .= "POST ".$url_info['path']." HTTP/1.1\n";
$request .= "Host: ".$url_info['host']."\n";
$request .= "Content-type: application/x-www-form-urlencoded\n";
$request .= "Content-length: ".strlen($data)."\n";
$request .= "Connection: close\n";
$request .= "\n";
$request .= $data."\n";
}

$fp = fsockopen($url_info["host"], $url_info["port"]);
fputs($fp, $request);//把HTTP头发送出去

$inheader = 1;
while(!feof($fp))
{
//$result 是提交后返回的数据
$result .= fgets($fp, 1024);
}
echo $result;
fclose($fp);
?>

回复讨论(解决方案)

41001 缺少access_token参数

返回码说明

我的url里面带access_token参数了啊，为上面会这样，求指导

$fp = fsockopen($url_info["host"], $url_info["port"]);
和
$request .= "POST ".$url_info['path']." HTTP/1.1\n";
$request .= "Host: ".$url_info['host']."\n";
$request .= "Content-type: application/x-www-form-urlencoded\n";
$request .= "Content-length: ".strlen($data)."\n";
$request .= "Connection: close\n";
$request .= "\n";
$request .= $data."\n";
里没有发现有token的信息。

access_token在url里面，get参数，这个参数应该怎么加

$request .= "POST ". $url_info['path']." HTTP/1.1\n";
这里填写带路径和参数的目标页名称，比如
/cgi-bin/qrcode/create?access_token=？？？？？
无论是 get 还是 post 方式，都是这样写

把参数token拼接在path后面的确可以了