PayPal AdaptivePayments PaymentDetail PayKey

Question

我正在使用 PayPal Pay API 和自适应（链式）付款。我正在尝试将用户转发到 PayPal，然后返回到我预定义的 return_url。

问题是：我的返回网址中需要有一个 PayKey。原因：我需要调用 PaymentDetail API 来查看 return_url 中的付款。而且，我不想使用 IPN，因为我需要在返回 URL 上使用一些令牌进行验证。

我遇到的问题是： PayKey 是使用所有参数生成的，包括 return-url （因此在我构建从中获取 $response 的实际数组之后。我无法将 PayKey 放在 return-Url 中因为此时尚未生成

  //Create request payload with minimum required parameters
  $bodyparams = array ("requestEnvelope.errorLanguage" => "en_US",
                       "actionType" => "PAY",
                       "currencyCode" => "USD",
                       "cancelUrl" => "http://www.paypal.com", 
                       "returnUrl" => $return_url . "&payKey=${payKey}",  **// Does not work - PAYKEY NEEDED TO ADD???**
                       "receiverList.receiver(0).email" => "[email protected]", //TODO
                       "receiverList.receiver(0).amount" => $price, //TODO
                       "receiverList.receiver(0).primary" => "true", //TODO
                       "receiverList.receiver(1).email" => "[email protected]", //TODO
                       "receiverList.receiver(1).amount" => $receiver_gets, //TODO
                       "receiverList.receiver(1).primary" => "false" //TODO
                       );

   // convert payload array into url encoded query string
   $body_data = http_build_query($bodyparams, "", chr(38));   // Generates body data

   try
   {
     //create request and add headers
     $params = array("http" => array(
                     "method" => "POST",
                     "content" => $body_data,
                     "header" =>  "X-PAYPAL-SECURITY-USERID: " . $API_UserName . "\r\n" .
                     "X-PAYPAL-SECURITY-SIGNATURE: " . $API_Signature . "\r\n" .
                     "X-PAYPAL-SECURITY-PASSWORD: " . $API_Password . "\r\n" .
                     "X-PAYPAL-APPLICATION-ID: " . $API_AppID . "\r\n" .
                     "X-PAYPAL-REQUEST-DATA-FORMAT: " . $API_RequestFormat . "\r\n" .
                     "X-PAYPAL-RESPONSE-DATA-FORMAT: " . $API_ResponseFormat . "\r\n"
                     ));

     //create stream context
     $ctx = stream_context_create($params);

     //open the stream and send request
     $fp = @fopen($url, "r", false, $ctx);

     //get response
     $response = stream_get_contents($fp);

     //check to see if stream is open
     if ($response === false) {
         throw new Exception("php error message = " . "$php_errormsg");
     }

     fclose($fp);

     //parse the ap key from the response
     $keyArray = explode("&", $response);

     foreach ($keyArray as $rVal){
       list($qKey, $qVal) = explode ("=", $rVal);
               $kArray[$qKey] = $qVal;
     }

     //set url to approve the transaction
     $payPalURL = "https://www.sandbox.paypal.com/webscr?cmd=_ap-payment&paykey=" . $kArray["payKey"]; **// Here it works fine, since the PayKey is generated at this point ...**

     //print the url to screen for testing purposes
     If ( $kArray["responseEnvelope.ack"] == "Success") {
       echo '<p><a href="' . $payPalURL . '" target="_blank">' . $payPalURL . '</a></p>';
      }
     else {
       echo 'ERROR Code: ' .  $kArray["error(0).errorId"] . " <br/>";
       echo 'ERROR Message: ' .  urldecode($kArray["error(0).message"]) . " <br/>";
     }

有人可以帮忙吗？

Answer 1

我自己也从事这个工作很多年了。我终于想通了。 Paypal 文档很难理解。我在下载的 paypal 自适应 pdf 指南中找到了答案。它指定将 payKey=${payKey} 添加到 return_url 的末尾。我刚刚在那里尝试过，对我的返回网址的 paypal 获取请求现在包含 paykey。

所以在 Rails 中我使用的 return_url 看起来像这样。按照指南的指示将 php 变量（我认为）写入 url

:return_url      => "http://***********.com/paypal-return?payKey=${payKey}"

Answer 2

您似乎遗漏了序列中的一个步骤。

第一步是将您的交易参数发送至

https://svcs.paypal.com/AdaptivePayments/Pay&yourtransactionparameters=blah
[sandbox]https://svcs.sandbox.paypal.com/AdaptivePayments/Pay&yourtransactionparameters=blah

您将在此响应中获得付款密钥。

成功检索到 paykey 后，您将调用：

https://www.paypal.com/webscr&cmd=_ap-payment&paykey=xxxx
[sandbox]https://www.sandbox.paypal.com/webscr&cmd=_ap-payment&paykey=xxxx

在第二次调用中，payKey 代表交易的其余部分，因此您不必构建另一个巨大的查询字符串。

Answer 3

更改代码：

//Create request payload with minimum required parameters
  $bodyparams = array ("requestEnvelope.errorLanguage" => "en_US",
                       "actionType" => "PAY",
                       "currencyCode" => "USD",
                       "cancelUrl" => "http://www.paypal.com", 
                       "returnUrl" => $return_url . '&payKey=${payKey}',  **// That's right**
                       "receiverList.receiver(0).email" => "[email protected]", //TODO
                       "receiverList.receiver(0).amount" => $price, //TODO
                       "receiverList.receiver(0).primary" => "true", //TODO
                       "receiverList.receiver(1).email" => "[email protected]", //TODO
                       "receiverList.receiver(1).amount" => $receiver_gets, //TODO
                       "receiverList.receiver(1).primary" => "false" //TODO
                       );

PHP 将 ${payKey} 解释为双引号之间的变量。
将双引号 (") 更改为单引号 (')

Answer 4

显然，您可以在返回 URL 中嵌入动态变量： https://www.x.com/thread/49785< /a>

不幸的是，{$payKey} 无效。所以它必须是 $paykey 或 $pay_key。祝你好运！