JSON RPC Java dzuvinov-如何设置用户名和密码

Question

我正在使用来自dzuvinov的JSON-RPC 2.0的Java库。我在为呼叫设置基本访问身份验证的用户名和密码时遇到问题。我的代码看起来像：

public static void main(String[] args)
{
    URL serverURL = null;
    try {
        serverURL = new URL("http://user:pass@127.0.0.1:18332/");
    } catch (MalformedURLException e) {
        System.err.println(e.getMessage());
        return;
    }
     JSONRPC2Session mySession = new JSONRPC2Session(serverURL);
     String method = "getinfo";
     int requestID = 0;
     JSONRPC2Request request = new JSONRPC2Request(method, requestID);
     JSONRPC2Response response = null;
     try {
             response = mySession.send(request);
     } catch (JSONRPC2SessionException e) {
             System.err.println(e.getMessage());
             return;
     }
     if (response.indicatesSuccess())
        System.out.println(response.getResult());
    else
        System.out.println(response.getError().getMessage());
}

我得到的回复是：

Network exception: Server returned HTTP response code: 401 for URL: http://user:pass@127.0.0.1:18332/

在Python中尝试类似的代码，我得到了正确的结果：

access = jsonrpc.ServiceProxy("http://user:pass@127.0.0.1:18332/")
print access.getinfo()

如何配置JSON RPC调用的基本访问身份验证？

Answer 1

final String rpcuser ="...";
final String rpcpassword ="...";
Authenticator.setDefault(new Authenticator() {
  protected PasswordAuthentication getPasswordAuthentication() {
      return new PasswordAuthentication (rpcuser, rpcpassword.toCharArray());
  }});

Answer 2

不确定"dzuvinov"中的库如何处理流。但您可以应用此代码在URL类打开的HTTP传输级别上执行基本身份验证。

URL url = new URL("http://user:pass@127.0.0.1:18332/");
URLConnection conn = url.openConnection();
String userpass = rpcuser + ":" + rpcpassword;
String basicAuth = "Basic " + new String(new Base64().encode(userpass.getBytes()));
conn.setRequestProperty ("Authorization", basicAuth);
conn.setRequestProperty ("Content-Type", "application/json");
conn.setDoOutput(true);

完整代码可在此处找到http://engnick.blogspot.ru/2013/03/java-trac-rpc-and-json.html。我用它来执行JSON-RPC的东西来与Trac的API通信。