Spring integration Java DSL http outbound

Question

How to resend same or modified message from outbound http call in case of specific client error responses like 400, 413 etc

@Bean
private IntegrationFlow myChannel() {
    IntegrationFlowBuilder builder = 
             IntegrationFlows.from(queue)
              .handle(//http post method config)
                    ...
                  .expectedResponseType(String.class))
              .channel(MessageChannels.publishSubscribe(channel2));
   return builder.get();
}

@Bean
private IntegrationFlow defaultErrorChannel() {
    
}

EDIT: Added end point to handle method

@Bean
private IntegrationFlow myChannel() {
    IntegrationFlowBuilder builder = 
             IntegrationFlows.from(queue)
              .handle(//http post method config)
                    ...
                  .expectedResponseType(String.class), 
                               e -> e.advice(myRetryAdvice()))
              .channel(MessageChannels.publishSubscribe(channel2));
   return builder.get();
}

@Bean
public Advice myRetryAdvice(){
 ... // set custom retry policy
}

Custom Retry policy:

class InternalServerExceptionClassifierRetryPolicy extends 
      ExceptionClassifierRetryPolicy {
    public InternalServerExceptionClassifierRetryPolicy() {
       final SimpleRetryPolicy simpleRetryPolicy = 
                  new SimpleRetryPolicy();
    simpleRetryPolicy.setMaxAttempts(2);

    this.setExceptionClassifier(new Classifier<Throwable, RetryPolicy>() {
        @Override
        public RetryPolicy classify(Throwable classifiable) {
            if (classifiable instanceof HttpServerErrorException) {
                // For specifically 500 and 504
                if (((HttpServerErrorException) classifiable).getStatusCode() == HttpStatus.INTERNAL_SERVER_ERROR
                        || ((HttpServerErrorException) classifiable)
                                .getStatusCode() == HttpStatus.GATEWAY_TIMEOUT) {
                    return simpleRetryPolicy;
                }
                return new NeverRetryPolicy();
            }
            return new NeverRetryPolicy();
        }
    }); 
}}

EDIT 2: Override open() to modify the original message

    RequestHandlerRetryAdvice retryAdvice = new 
    RequestHandlerRetryAdvice(){
    @Override
    public<T, E extends Throwable> boolean open(RetryContext 
    retryContext, RetryCallback<T,E> callback){
    Message<String> originalMsg = 
(Message)  retryContext.getAttribute(ErrorMessageUtils.FAILED_MESSAGE_CONTEXT); 
   Message<String> updatedMsg = //some updated message

retryContext.setAttribute(ErrorMessageUtils.FAILED_MESSAGE_CONTEXT,up datedMsg); return super.open(retryContext, callback); }

Answer 1

See a RequestHandlerRetryAdvice: https://docs.spring.io/spring-integration/reference/html/messaging-endpoints.html#message-handler-advice-chain. So, you configure some RetryPolicy to check those HttpClientErrorException for retry and the framework will re-send for you.

Java DSL allows us to configure it via second handle() argument - endpoint configurer: .handle(..., e -> e.advice(myRetryAdvice)): https://docs.spring.io/spring-integration/reference/html/dsl.html#java-dsl-endpoints