JAXB Marshaller initialization

Question

I need to Marshal a JAXB object to xml format string. I'm using a SLSB and hook the code to create the Marshaller and other things in @PostConstruct annotated method. So that each time I need not load the schema and create the Marshaller.

The code in @PostConstruct annotated method is as below.

  JAXBContext jaxbContext = JAXBContext.newInstance(jaxbPackageName);

  SchemaFactory factory = SchemaFactory
      .newInstance(javax.xml.XMLConstants.W3C_XML_SCHEMA_NS_URI);
  URL schemaUrl = Thread.currentThread().getContextClassLoader()
      .getResource(resourcePath);
  schema = factory.newSchema(schemaUrl);

  setMarshaller(jaxbContext.createMarshaller());

  getMarshaller().setProperty(Marshaller.JAXB_FORMATTED_OUTPUT,
      Boolean.FALSE);
  getMarshaller().setSchema(schema);     

  getMarshaller().setEventHandler(new DefaultValidationEventHandler());

  setUnmarshaller(jaxbContext.createUnmarshaller());
  getUnmarshaller().setSchema(schema);
  getUnmarshaller().setEventHandler(new DefaultValidationEventHandler());

And when a client code needs xml format of the object the following method returns the same.

    OutputStream outputStream = new ByteArrayOutputStream();
   getMarshaller().setProperty(Marshaller.JAXB_SCHEMA_LOCATION,
        schemaLocation);
    getMarshaller().marshal(document, outputStream);
    xmlString = outputStream.toString();

My concern is, is there any better way to do the same (whenever a client code wants to get xml format of a JAXB object, the fastest approach to return the same? ).

Thanks

Valchev Valchev · Accepted Answer · 2011-12-19T14:30:45

I don't know any other way of marshalling/unmarshalling with JAXB. When I had the same problem, I checked the CXF code and it was the same.

Refer to JAXB Performance and Thread safety

JAXB Marshaller initialization

2 Answers