Pinterest API中的Auth异常

Question

我得到{“状态”：“失败”，“代码”：3，“主持人”：“coreapp-devplatform-devapi-173”，“generated_at”：“星期二，2015年9月29日05:42:42 +0000“，”message“：”授权失败。“，”data“：null}异常，同时调用https://api.pinterest.com/v1/oauth/token?code= &client_id=&grant_type=authorization_code. Can anyone help me to resolve this?

Code:

            if (request.getParameter("code") != null) {
                code = request.getParameter("code");
            }
            String authorizeUrl = "https://api.pinterest.com/v1/oauth/token";
            String postStr = "code=" + code + "&client_id=" + clientId + "&grant_type=authorization_code";
            String responseStr = postTokenRequest(authorizeUrl, postStr);

//Method
public String postTokenRequest(String serverURL, String content) {
            String inputLine = "";
            HttpsURLConnection connection = null;
            URL url = new URL(serverURL);
            String method = "POST";
            InputStream connectionIn = null;
            BufferedReader buffer = null;
            try {
                connection = (HttpsURLConnection) url.openConnection();
                connection.setRequestProperty("Content-Length", String.valueOf(content.getBytes().length));
                connection.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
                connection.setDoOutput(true);
                connection.getOutputStream().write(content.getBytes());
                int returnCode = connection.getResponseCode();
                if (returnCode == 200) {
                    connectionIn = connection.getInputStream();
                } else {
                    connectionIn = connection.getErrorStream();
                }
                buffer = new BufferedReader(new InputStreamReader(connectionIn));
                StringBuilder sb = new StringBuilder();
                while ((inputLine = buffer.readLine()) != null) {
                    sb.append(inputLine);
                }
                buffer.close();
                return sb.toString();
      }catch(Exception e){}
  }