Question

Ok, I’ve searched everywhere for an answer to this. It is driving me nuts.

All I need to do is unmarshal a very simple webservice response. The only problem is, I am using a generated source file without the @XmlRootElement annotation. I am unable to edit this generated source file to add @XmlRootElement, either. I need to use it “as is”.

This is the current code that I have, but it is resulting in an error shown at the bottom of this post. I have tried to use a JAXBElement wrapper but to no avail. Could somebody please give me the code I need? I have no idea how to use "QName"s etc.

This code below works great with classes that have @XmlRootElement:

 MyGeneratedClass response = restTemplate.getForObject("url to webservice!"),
     MyGeneratedClass.class);

 return response

Sadly, it is producing this error in this case. Please help me to unmarshal the REST response!

 Could not extract response: no suitable HttpMessageConverter found for response
    type [MyGeneratedClass] and content type [application/xml;version=1]

Answer 1

I forgot about posting this many months ago and I should probably follow it up with the solution.
This solution also adds a cookie to the request headers, but you can ignore that.
In the case that a generated source file does not have @XmlRootElement annotation, you can unmarshal as follows:

// Cookie setting
HttpHeaders requestHeaders = new HttpHeaders();
requestHeaders.set("Cookie", "myCookie=value");
HttpEntity<?> requestEntity = new HttpEntity(requestHeaders);

HttpEntity<String> response = restTemplate.exchange("web service url"), 
    HttpMethod.GET, requestEntity, String.class);

// Unmarshalling
JAXBElement<MyGeneratedClass> result = 
    (JAXBElement<MyGeneratedClass>) unmarshaller.unmarshal(
        new StreamSource(new ByteArrayInputStream(response.getBody().getBytes())));

return result.getValue();