在Apache CXF Interceptor中编写消息内容和响应代码

Question

我试图通过使其中一个方法需要HTTP基本身份验证来确保我的Web服务安全.为了做到这一点,我实现了一个自定义的Interceptor调用LoginInterceptor,它检查请求的URL,如果它对应于一个被调用的方法open,它会检查消息头是否有用户名和密码.

如果标头中没有验证字段,则响应代码设置为HTTP_UNAUTHORIZED,如果用户名或密码不正确,则将响应代码设置为HTTP_FORBIDDEN.这是代码:

public LoginInterceptor() {
     super(Phase.RECEIVE);
     addAfter(RequestInterceptor.class.getName()); //another custom interceptor, for some simple redirections.
}

public void handleMessage(Message message) throws Fault {
     String requestURI = message.get(Message.REQUEST_URI).toString();
     String methodKeyword = requestURI.substring("localhost".length()+1).split("/")[0];

     if(methodKeyword.equals("open")) {
          AuthorizationPolicy policy = message.get(AuthorizationPolicy.class);
          if(policy == null) {
              sendErrorResponse(message, HttpURLConnection.HTTP_UNAUTHORIZED);
              return;
          }

          //userPasswords is a hashmap of usernames and passwords.     
          String realPassword = userPasswords.get(policy.getUserName());
          if (realPassword == null || !realPassword.equals(policy.getPassword())) {
                    sendErrorResponse(message, HttpURLConnection.HTTP_FORBIDDEN);
          }
     }
}

//This is where the response code is set, and this is where I'd like to write the response message.
private void sendErrorResponse(Message message, int responseCode) {
    Message outMessage = getOutMessage(message);
    outMessage.put(Message.RESPONSE_CODE, responseCode);

    // Set the response headers
    Map responseHeaders = (Map) message.get(Message.PROTOCOL_HEADERS);

    if (responseHeaders != null) {
        responseHeaders.put("Content-Type", Arrays.asList("text/html"));
        responseHeaders.put("Content-Length", Arrays.asList(String.valueOf("0")));
    }

    message.getInterceptorChain().abort();

    try {
        getConduit(message).prepare(outMessage);
        close(outMessage);
    } catch (IOException e) {
        e.printStackTrace();
    }
}

private Message getOutMessage(Message inMessage) {
    Exchange exchange = inMessage.getExchange();
    Message outMessage = exchange.getOutMessage();

    if (outMessage == null) {
            Endpoint endpoint = exchange.get(Endpoint.class);
            outMessage = endpoint.getBinding().createMessage();
            exchange.setOutMessage(outMessage);
    }

    outMessage.putAll(inMessage);

    return outMessage;

}

//Not actually sure what this does. Copied from a tutorial online. Any explanation is welcome
private Conduit getConduit(Message inMessage) throws IOException {
    Exchange exchange = inMessage.getExchange();
    Conduit conduit = exchange.getDestination().getBackChannel(inMessage);
    exchange.setConduit(conduit);
    return conduit;
}

private void close(Message outMessage) throws IOException {
    OutputStream os = outMessage.getContent(OutputStream.class);
    os.flush();
    os.close();
}

这工作正常,但是,我想在响应中返回一条消息,例如"用户名或密码不正确".我已尝试从sendErrorResponse方法中做:

outMessage.setContent(String.class, "incorrect username or password") 我将内容长度设置为"incorrect username or password".length().这不起作用,我想因为Apache CXF消息使用InputStreams和OutputStreams.

所以我尝试过:

OutputStream os = outMessage.getContent(OutputStream.class);
try {
    os.write("incorrect username or password".getBytes() );
    outMessage.setContent(OutputStream.class, os);
 } catch (IOException e) {
    e.printStackTrace();
 }

这也不起作用.当用调试器逐句通过,我注意到,os就是null当邮递员测试,我得到:

无法获得任何响应这似乎是连接到的错误http://localhost:9090/launcher/open.响应状态为0.有关何时发生此更多详细信息,请查看W3C XMLHttpRequest Level 2规范.

在Chrome中按ctrl + shif + c(打开开发工具),然后检查网络标签,我看到:

"ERR_CONTENT_LENGTH_MISMATCH"

我已经尝试过使用XMLStreamWriter,但是没有更好的.

问题:

• 我可以返回正确的响应代码(401 Unauthorized和403 forbidden),但是如何在响应正文中返回消息？

• 我是否需要专门扩展特定的OutInterceptor,如JASRXOutInterceptor才能修改消息内容？

• 我之前尝试过使用JAASInterceptor,但我没有设法让它工作.有人可以告诉我如何以这种方式实现它,如果这在某种程度上更容易吗？

• 我也可以像这样抛出一个错误:throw new Fault("incorrect username or password", Logger.getGlobal());但是HTTP响应代码将是500.我宁愿返回正确的401或403响应.

注意:

现在我仍在使用HTTP作为传输层.解决这个问题后,我将更改为HTTPS.

Answer 1

基本上,我想要做的是返回错误,HTTP响应代码为401(未授权)或403(禁止)而不是500(服务器错误).事实证明Apache CXF提供了一种简单的方法,使用Fault.setStatusCode(int) method,正如我从Stack Overflow上的这个问题中发现的:如何在Apache CXF中抛出403错误 - Java

所以这就是我的handleMessage方法现在的样子:

public void handleMessage(Message message) throws Fault {
        String requestURI = message.get(Message.REQUEST_URI).toString();
        String methodKeyword = requestURI.substring("localhost".length()+1).split("/")[0];

        if(methodKeyword.equals("open")) {
            AuthorizationPolicy policy = message.get(AuthorizationPolicy.class);

            if(policy == null) {
                Fault fault = new Fault("incorrect username or password", Logger.getGlobal());
                fault.setStatusCode(401);
                throw fault;
            }

            String realPassword = userPasswords.get(policy.getUserName());
            if (realPassword == null || !realPassword.equals(policy.getPassword())) {
                Fault fault = new Fault("incorrect username or password", Logger.getGlobal());
                fault.setStatusCode(403);
                throw fault;
            }
        }
    }

我删除了其他方法,它们是不必要的.