Как преобразовать нижеприведенную конфигурацию XML интеграции Spring в bean-компоненты конфигурации Java.
<?xml version="1.0" encoding="UTF-8"?> <beans xmlns="http://www.springframework.org/schema/beans" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:int="http://www.springframework.org/schema/integration" xmlns:int-http="http://www.springframework.org/schema/integration/http" xsi:schemaLocation=" http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd http://www.springframework.org/schema/integration http://www.springframework.org/schema/integration/spring-integration.xsd http://www.springframework.org/schema/integration/http http://www.springframework.org/schema/integration/http/spring-integration-http.xsd"> <int:channel id='reply.channel'> <int:queue capacity='10' /> </int:channel> <int:channel id='get.request.channel'/> <int-http:outbound-gateway id="outbound.gateway" request-channel="get.request.channel" url="{fhirurl}" http-method-expression="payload.getHttpMethod()" expected-response-type-expression="payload.getResponseType()" charset="UTF-8" reply-timeout="5000" reply-channel="reply.channel"> <int-http:uri-variable name="fhirurl" expression="payload.getUrl()"/> </int-http:outbound-gateway> </beans>
Я получил код ниже, но не могу его использовать. Можете ли вы мне помочь, что написать в soapHeaderMapper и requestFactory
@Bean public IntegrationFlow httpOut() { return IntegrationFlows.from("reply.channel") .handle(Http.outboundGateway("http://localhost:8080/") .charset("UTF-8") .httpMethod(HttpMethod.GET) .headerMapper(soapHeaderMapper()) .requestFactory(requestFactory()), e -> e.id("test")) .get(); }
Заранее спасибо