Parse Php推送通知返回true。如何防止它返回响应

我在Android和iOS上成功设置了解析推送通知。我使用PHP根据设备的用户ID向设备发送推送通知。推送通知正在成功传递。然而，在运行PHP代码时，parse似乎返回true的响应。有没有办法防止它“呼应”响应

PHP代码-

<?php

$APPLICATION_ID = "XXXX";
$REST_API_KEY = "XXXX";

$url = 'https://api.parse.com/1/push';
$data = array(
    'where' => array(
    'user_id' => array('$in' => array('80','19','20','21'))
    ),
    'data' => array(
        'alert' => 'greetings programs',
    ),
);
$_data = json_encode($data);
$headers = array(
    'X-Parse-Application-Id: ' . $APPLICATION_ID,
    'X-Parse-REST-API-Key: ' . $REST_API_KEY,
    'Content-Type: application/json',
    'Content-Length: ' . strlen($_data),
);

$curl = curl_init($url);
curl_setopt($curl, CURLOPT_POST, 1);
curl_setopt($curl, CURLOPT_POSTFIELDS, $_data);
curl_setopt($curl, CURLOPT_HTTPHEADER, $headers);
curl_exec($curl);

?>

答复：

{“结果”：真}

---

有没有办法防止它重复响应-{“result”：true}

使用-curl\u setopt（$curl，CURLOPT\u RETURNTRANSFER，1）

以下是ref的完整PHP Rest API CURL代码：

<?php

$APPLICATION_ID = "xxxx";
$REST_API_KEY = "xxxx";

$url = 'https://api.parse.com/1/push';

$data = array(
    'where' => array(
        'objectId' => array('$in' => array('xx'))
    ),
    'data' => array(
        'alert' => 'test parse message via php script',
    ),
);

$_data = json_encode($data);

$headers = array(
    'X-Parse-Application-Id: ' . $APPLICATION_ID,
    'X-Parse-REST-API-Key: ' . $REST_API_KEY,
    'Content-Type: application/json',
    'Content-Length: ' . strlen($_data),
);

$curl = curl_init($url);
curl_setopt($curl, CURLOPT_POST, 1);
curl_setopt($curl, CURLOPT_POSTFIELDS, $_data);
curl_setopt($curl, CURLOPT_HTTPHEADER, $headers);
curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);

$result = curl_exec($curl);
echo $result;
print_r($result);
die('finished!');
?>

注意：它只在我的服务器GoDaddy上工作，不在本地主机上工作-不知道为什么。可能是godaddy版本高于5.4.x