AWS Lambda Java集成

Question

我想将我的Saml应用程序与AWS集成起来，但不幸的是，我的Saml代码接受了它的输入，如下所示。我需要发送HttpServletRequest和HttpServletResponse作为我的java处理程序的输入。因此，它需要request和response作为输入，但我的lambda处理程序只将输入作为JSON或java，而我对如何继续操作感到困惑。

    public Authentication attemptAuthentication(HttpServletRequest request, HttpServletResponse response) throws AuthenticationException {
    //validation 
    //output
  return authentication;
    }

Answer 1

AWS团队创建了一个无服务器包装器，用于公开请求和响应对象。这应该允许您执行您的need.In处理程序，您可以实现一个新接口，并且它们的底层功能会将请求和响应作为AwsProxyRequest和AwsProxyResponse返回给您，后者应该是HttpServletRequest和HttpServletResponse的子级。

码

public class StreamLambdaHandler implements RequestStreamHandler {
    private SpringLambdaContainerHandler<AwsProxyRequest, AwsProxyResponse> handler;
    private Logger log = LoggerFactory.getLogger(StreamLambdaHandler.class);

    @Override
    public void handleRequest(InputStream inputStream, OutputStream outputStream, Context context)
            throws IOException {
        if (handler == null) {
            try {
                handler = SpringLambdaContainerHandler.getAwsProxyHandler(PetStoreSpringAppConfig.class);
            } catch (ContainerInitializationException e) {
                log.error("Cannot initialize Spring container", e);
                outputStream.close();
                throw new RuntimeException(e);
            }
        }

        AwsProxyRequest request = LambdaContainerHandler.getObjectMapper().readValue(inputStream, AwsProxyRequest.class);

        AwsProxyResponse resp = handler.proxy(request, context);

        LambdaContainerHandler.getObjectMapper().writeValue(outputStream, resp);
        // just in case it wasn't closed by the mapper
        outputStream.close();
    }
}

源-> https://github.com/awslabs/aws-serverless-java-container