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 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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