Hi Friends, I am working on a small spike in which my task is to pick an xml from a directory, put it on jms queue and save the xml to some other directory.
This is the code snapshot, this is after I converted the file contents in to a string. private static final String URI = "activemq://queue:test.MyQueue" ; protected void sendTo(String uri,String message) throws Exception { ProducerTemplate producerTemplate = camelContext.createProducerTemplate(); Exchange exchange = new DefaultExchange(camelContext, ExchangePattern.InOut); exchange.getIn().setBody(message); Message camelMessage = exchange.getIn(); camelMessage.setHeader(Exchange.FILE_NAME, "testmesage"); System.out.println(exchange.getExchangeId()); System.out.println(exchange.getIn().getMessageId()); producerTemplate.send(uri, exchange); Thread.sleep(1000); } and camel route is <route> <from uri="jms:test.MyQueue"/> <to uri="file://target/test?noop=true/> </route> The file appears in target but with name ID-localhost-46447-1346755756577-0-2-2-1-1 and not "testmessage" as I specified. hope I am clear Please help Thanks in advance Saurabh -- View this message in context: http://camel.465427.n5.nabble.com/Naming-your-file-with-org-apache-camel-file-name-tp473074p5718602.html Sent from the Camel - Users mailing list archive at Nabble.com.