How to omit fields when using camel-xmljson JAR to convert XML string to JSON string - json

Question

XmlJsonDataFormat xmlJsonFormat = new XmlJsonDataFormat();
xmlJsonFormat.setEncoding("UTF-8");
xmlJsonFormat.setForceTopLevelObject(true);
xmlJsonFormat.setTrimSpaces(true);
xmlJsonFormat.setSkipNamespaces(true);
xmlJsonFormat.setRemoveNamespacePrefixes(true);
Pool a XML file from source directory and transform it to JSON format before sending it to destination directory
from("file:/inn?noop=true").marshal(xmlJsonFormat)
.setHeader(Exchange.FILE_NAME, simple("${file:onlyname.noext}.json"))
.to("file:/out").end();
It works great when i want a JSON string right from a XML without using a POJO.
But now consider a XML where I would like to omit an element.
Should I use another lib but camel-xmljson?
Is this way the right way of accomplish it:
JAXB lib to get a POJO from XML
Jackson lib to get a JSON from POJO and with excluding POJO fields from marshalling

Answer 1

I understand, even I would try to avoid any POJO's in camel routes.
Having said that, I wont just straight way let the XML out. I would put a XSLT after the xmljson component, so that I can trim the xml for desired output.

Answer 2

Thanks gnanagurus! This is quite simple using XSLT. Moreover the xslt: component is included in the last version of the core of Camel.
I defined an xlst file 'transform.xsl' taht includes the field that is to be omited. And then defined the Camel route like that:
from("file:/inn?noop=true")
.to("xslt:file:src/transform.xsl")
.marshal(xmlJsonFormat)
.setHeader(Exchange.FILE_NAME, simple("${file:onlyname.noext}.json"))
.to("file:/out").end();