Hi All, While working i came across a requirements where i have to name the final output file based on the input file e.g if the name of the input file is a.xml and i need to convert it to csv at the end than the output file should be named as customoutput_a.xml.csv like in my case i have the following route
<route id="demo"> <from uri="{{demo.filePath}}"/> <choice> <when> <simple>${file:ext} == 'csv'</simple> <unmarshal> <csv/> </unmarshal> <to uri="bean:converter?method=process" /> <to uri="file://data/output?fileName=tmpXML.xml" /> </when> </choice> <to uri="{{demo.tmpXMLtoStandardXML}}"/> <to uri="file://data/output?fileName=3_StandardXML.xml" /> <to uri="{{demo.StandardXMLToImpexXslFile}}"/> <to uri="{{demo.targetXMLFile}}"/> </route> now in such case in the last route point i have to name the file based on the input file name, can any one help me how i can pass the input file name throughout the route in camel so that i can extract the file name when producing final output Thanks in advance -- View this message in context: http://camel.465427.n5.nabble.com/Passing-input-fileName-along-the-complete-route-tp4733437p4733437.html Sent from the Camel - Users mailing list archive at Nabble.com.