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将php curl转换为iTunes InApp verifyReceipt的GAE urlfetch

发布时间:2020-12-13 22:52:02 所属栏目:PHP教程 来源:网络整理
导读:有人可以帮助将这个 PHP Curl转换为UrlFetch吗?这用于Apple iTunes verifyReceipt if (getiTunesProductionLevel($app_id)=="sandbox" || $sandbox_override == TRUE) { $endpoint = 'https://sandbox.itunes.apple.com/verifyReceipt';}else { $endpoint =
有人可以帮助将这个 PHP Curl转换为UrlFetch吗?这用于Apple iTunes verifyReceipt

if (getiTunesProductionLevel($app_id)=="sandbox" || $sandbox_override == TRUE) {
    $endpoint = 'https://sandbox.itunes.apple.com/verifyReceipt';
}
else {
    $endpoint = 'https://buy.itunes.apple.com/verifyReceipt';
}

$postData = json_encode(array(
    'receipt-data' => $receipt,'password' => $sharedSecret));

$ch = curl_init($endpoint);
curl_setopt($ch,CURLOPT_SSL_VERIFYHOST,0);
curl_setopt($ch,CURLOPT_SSL_VERIFYPEER,CURLOPT_RETURNTRANSFER,true);
curl_setopt($ch,CURLOPT_POST,CURLOPT_POSTFIELDS,$postData);
$response = curl_exec($ch);
$errno = curl_errno($ch);
$errmsg = curl_error($ch);
curl_close($ch);

这是我能得到的最好的.但还不够好.

logMessage(LogType::Info,"XXX URLFetch 0");
    $postData = json_encode(array(
    'receipt-data' => $receipt,'password' => $sharedSecret));
$post_data = json_decode($postData);
logMessage(LogType::Info,"XXX URLFetch 1");
$data = http_build_query($post_data);
logMessage(LogType::Info,"XXX URLFetch 2");

$context = [
  'http' => [
    'method' => 'post','header' => "Content-Type: application/x-www-form-urlencodedrn",'content' => $data
  ]
];

logMessage(LogType::Info,"XXX URLFetch 3");
$context = stream_context_create($context);
logMessage(LogType::Info,"XXX URLFetch 4");
$result = file_get_contents($endpoint,false,$context);
logMessage(LogType::Info,"XXX result:" . $result);
$response = $result;
$errno = 0;
logMessage(LogType::Info,"XXX response:");

它能够发布但返回此响应
????XXX结果:{“status”:21002}

解决方法

你为什么在你的urlfetch代码中使用json_decode $post_data而不是curl?

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