Question

我正在研究spring integration dsl。要求是从队列中读取xml消息，根据消息头值，我需要调用不同的服务。我能够从队列中获取消息，但无法在dsl中编写代码以将xml消息解组为object。有人可以帮忙吗？我有我的unmarshaller但无法用dsl

连线

 IntegrationFlows
            .from(Jms.inboundGateway(connectionFactory)
                    .destination(someQueue)
                    .configureListenerContainer(spec -> spec.get().setSessionTransacted(true)))
            .transform(??)

Answer 1

首先，您可以使用Jms.inboundGateway()配置MarshallingMessageConverter：

/**
 * @param messageConverter the messageConverter.
 * @return the spec.
 * @see ChannelPublishingJmsMessageListener#setMessageConverter(MessageConverter)
 */
public S jmsMessageConverter(MessageConverter messageConverter) {
    this.target.getListener().setMessageConverter(messageConverter);
    return _this();
}

但如果您仍坚持使用.transform()，请考虑使用UnmarshallingTransformer：

/**
 * An implementation of {@link Transformer} that delegates to an OXM
 * {@link Unmarshaller}. Expects the payload to be of type {@link Document},
 * {@link String}, {@link File}, {@link Source} or to have an instance of
 * {@link SourceFactory} that can convert to a {@link Source}. If
 * {@link #alwaysUseSourceFactory} is set to true, then the {@link SourceFactory}
 * will be used to create the {@link Source} regardless of payload type.
 * <p>
 * The {@link #alwaysUseSourceFactory} is ignored if payload is
 * {@link org.springframework.ws.mime.MimeMessage}.
 * <p>
 * The Unmarshaller may return a Message, but if the return value is not
 * already a Message instance, a new Message will be created with that
 * return value as its payload.
 *
 * @author Jonas Partner
 * @author Artem Bilan
 */
public class UnmarshallingTransformer extends AbstractPayloadTransformer<Object, Object> {

https://docs.spring.io/spring-integration/docs/5.0.4.RELEASE/reference/html/xml.html#xml-unmarshalling-transformer

如何使用spring integration dsl解组xml

1 个答案: