How to force persisting data during execution of camel processors rather than the end of the route? - spring-boot

Question

When i toogle a breakpoint beside the repository.saveAndFlush and during de debug mode i see that it returne a new client objet with new Id but when i check in the data base i do not find that client. However, if i do a resume (F8 with eclipse) then i re-check the DB i find the client.
So How to force persisting data during execution of camel processors rather than the end of the route?
#Component
public class myRoute extends RouteBuilder {
#Autowired Processor validationDatasProcessor;
#Autowired Processor clientProcessor;
#Autowired Processor endCientProcessor;
#Override
public void configure() throws Exception {
from("queueIn")
.id("route_processing").messageHistory().transacted()
.log(LoggingLevel.DEBUG, log, "reception").pipeline()
.process(validationDatasProcessor)
.id(validationDatasProcessor.getClass().getSimpleName().toLowerCase())
.process(clientProcessor)
.id(clientProcessor.getClass().getSimpleName().toLowerCase())
.process(endCientProcessor).id(endCientProcessor.getClass().getSimpleName().toLowerCase())
.to("outputQueue")
.end();
}
}
Processors:
#Component
public class ValidationDatasProcessor implements Processor {
#Autowired ObjectMapper objectMapper;
#Autowired ClientRepository clientRepository;
#Override
public void process(Exchange exchange) throws Exception {
String clientString = exchange.getIn().getBody(String.class);
Client client = objectMapper.readValue(clientString, Client.class);
clientRepository.saveAndFlush(client)
exchange.setOut(generateOutMessage(client, exchange.getContext()));
}
Message generateOutMessage(Client client, CamelContext camelContext) throws JsonProcessingException {
DefaultMessage outMessage = new DefaultMessage(camelContext);
outMessage.setBody(objectMapper.writeValueAsString(client), String.class);
return outMessage;
}
}
#Component
public class ClientProcessor implements Processor {
#Autowired ObjectMapper objectMapper;
#Autowired ClientRepository clientRepository;
....
#Override
public void process(Exchange exchange) throws Exception {
String clientString = exchange.getIn().getBody(String.class);
Client client = objectMapper.readValue(clientString, Client.class);
client.setAccessDate(LocalDateTime.now);
clientRepository.saveAndFlush(client)
exchange.setOut(generateOutMessage(client, exchange.getContext()));
}
Message generateOutMessage(Client client, CamelContext camelContext) throws JsonProcessingException {
DefaultMessage outMessage = new DefaultMessage(camelContext);
outMessage.setBody(objectMapper.writeValueAsString(client), String.class);
return outMessage;
}
}

Answer 1

Your entire route is a transacted; which means the whole route is under a transaction scope. A commit will be done only after the whole route is executed.
if you want to execute a processor outside the transaction boundary, split the route and use a seda endpoint. seda are asynchronous and start new threads. They won't participate in the active transaction boundary.
Commiting parts in the middle of a transaction scope doesn't sound like a great idea. Perhaps your rounte needs to be split into multiple fragments.
This documentation might help you understand them better.