3

我正在尝试创建简单的 PayPal Pay API 操作，当我从控制台运行此代码时，它给了我响应，即创建了付款。

现在，当我试图从我的控制器运行它时，它给了我

    Authentication+failed.+API+credentials+are+incorrect.

这是我的控制器：

     def pay
require 'httpclient'
require 'xmlsimple'
clnt = HTTPClient.new
credentials = {
    'USER' => 'payer_1342623102_biz_api1.gmail.com',
   'PWD' => '1342623141',
   'SIGNATURE' => 'Ay2zwWYEoiRoHTTVv365EK8U1lNzAESedJw09MPnj0SEIENMKd6jvnKL '
 }

header =  {"X-PAYPAL-SECURITY-USERID" => "payer_1342623102_biz_api1.gmail.com",
               "X-PAYPAL-SECURITY-PASSWORD" => "1342623141",
               "X-PAYPAL-SECURITY-SIGNATURE" => "Ay2zwWYEoiRoHTTVv365EK8U1lNzAESedJw09MPnj0SEIENMKd6jvnKL ",
               "X-PAYPAL-REQUEST-DATA-FORMAT" => "NV",
               "X-PAYPAL-RESPONSE-DATA-FORMAT" => "XML",
               "X-PAYPAL-APPLICATION-ID" =>  "APP-80W284485P519543T"
                }
data = {"actionType" => "PAY",
           "receiverList.receiver(0).email"=> "denmed_1342605975_biz@gmail.com",
           "receiverList.receiver(0).amount" => "10",
           "currencyCode" => "USD",
           "cancelUrl" => "http://127.0.0.1:3000",
           "returnUrl" => "http://127.0.0.1:3000",
           "requestEnvelope.errorLanguage" => "en_US"}
uri = "https://svcs.sandbox.paypal.com/AdaptivePayments/Pay"
res = clnt.post(uri, data, header)
@xml = XmlSimple.xml_in(res.content)
payKey = @xml["payKey"].to_s()
payKey = payKey.tr("[]", "")
payKey = payKey[1..20]
redirect_to "https://svcs.sandbox.paypal.com/AdaptivePayments/Pay?cmd=_ap-payment&paykey=#{payKey}"
end

一切都顺利吗 ？谁能提出我的请求失败的原因？

4

2 回答 2

2

一位好人发现了我的错误。我将用户重定向到错误的网址。

这一行：

redirect_to "https://svcs.sandbox.paypal.com/AdaptivePayments/Pay?cmd=_ap-payment&paykey=#{payKey}"

应该：

redirect_to "https://sandbox.paypal.com/webscr?cmd=_ap-payment&paykey=#{paykey}"
于 2012-07-19T06:15:22.430 回答
0

我得到了同样的错误：我意识到我忘了包括

sandbox_email_address: xxx@example.com

在我的 yml 文件中

于 2013-09-05T22:43:44.240 回答