2015-02-10 75 views
0

我试图创建一个Java相当于此:卷曲后在Java中

curl -X POST https://connect.stripe.com/oauth/token \ 
    -d client_secret=sk_test_px1LcMW6VWVSWKrRT4MuHvTY \ 
    -d code=AUTHORIZATION_CODE \ 
    -d grant_type=authorization_code 
在Java

按照这里找到DOC:https://stripe.com/docs/connect/oauth#token-request。这是我所到达的,但我一直未经授权。我是否设置了错误?我可以让请求正确地使用curl工作,但不能与java一起工作。

 HttpClient client = HttpClientBuilder.create().build(); 
     HttpPost post = new HttpPost("https://connect.stripe.com/oauth/token"); 
     try 
     { 
      List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1); 
      nameValuePairs.add(new BasicNameValuePair("grant_type", "authorization_code")); 
      nameValuePairs.add(new BasicNameValuePair("code", code)); 
      nameValuePairs.add(new BasicNameValuePair("client_secret", stripeClientId)); 
      post.setEntity(new UrlEncodedFormEntity(nameValuePairs)); 

      HttpResponse response = client.execute(post); 
      BufferedReader rd = new BufferedReader(new InputStreamReader(response.getEntity().getContent())); 
      String line = ""; 
      while ((line = rd.readLine()) != null) 
      { 
       System.out.println(line); 
      } 

     } 
     catch (IOException e) 
     { 
      e.printStackTrace(); 
     } 

回答

0

我想通了。看起来我正在用错误的键发出请求。这似乎是发送此请求的正确方法。