Lets say I have JSON of the following format:
{
\"type\" : \"Foo\"
\"data\" : {
\"object\" : {
\"id\" : \"1\"
\"fizz\
I think it is rather straight-forward. You probably have a super class that has properties for metadata
and owner
, so rather than making it truly generic, you could substitute T for your super class. But basically, you will have to parse the name of the class from the actual JSON string, which in your example would look something like this:
int start = jsonString.indexOf("type");
int end = jsonString.indexOf("data");
Class actualClass = Class.forName(jsonString.substring(start + 4, end - 2)); // that of course, is approximate - based on how you format JSON
and overall code could be something like this:
public static T deserialize(String xml, Object obj)
throws JAXBException {
T result = null;
try {
int start = jsonString.indexOf("type");
int end = jsonString.indexOf("data");
Class actualClass = Class.forName(jsonString.substring(start + 4, end - 2));
JAXBContextFactory factory = JAXBContextFactory.getInstance();
JAXBContext jaxbContext = factory.getJaxBContext(actualClass);
Unmarshaller jaxbUnmarshaller = jaxbContext.createUnmarshaller();
// this will create Java object
try (StringReader reader = new StringReader(xml)) {
result = (T) jaxbUnmarshaller.unmarshal(reader);
}
} catch (JAXBException e) {
log.error(String
.format("Exception while deserialising the object[JAXBException] %s\n\r%s",
e.getMessage()));
}
return result;
}