Converting a wrapper object to a value object

Question

I have ChannelWrapper and Channel Value Object. I want to convert each and every field of ChannelWrapper to Channel Value Object.

public class ChannelWrapper extends BaseWrapper implements APIWrapper<Channel>,
 APIUnwrapper<Channel> {
 @XmlElement
 private Long id;
 @XmlElement
 private User user;
 @XmlElement
 private Tenant tenant;
// Getter Setters
 @Override
 public Channel unwrap(HttpServletRequest arg0, ApplicationContext arg1) {
 // TODO Auto-generated method stub
 return null;
 }
 @Override
 public void wrapDetails(Channel model, HttpServletRequest request) {
 this.id = model.getId();
 this.user = model.getUser();
 this.tenant = model.getTenant();
 }
 @Override
 public void wrapSummary(Channel model, HttpServletRequest request) {
 wrapDetails(model, request);
}

Channel Value Object:

public class Channel {
 private Long id;
 private User user;
 private Tenant tenant;
 //getter setters }

Converter:

Channel channel = new Channel();
channel.setId(channelWrapper.getId());
channel.setUser(channelWrapper.getUser());
channel.setTenant(tenant);

Is this the best way to create a converter?

Any other general comments are appreciated.

• Gerold Broser: I'm sorry but I don't get your point. You have a wrapper, a.k.a. Adapter, for a Channel, ChannelWrapper. Now you'd like to wrap this wrapper into another wrapper (or converter as you call it) such leading to a ChannelWrapperWrapper or ChannelWrapperAdapter. And you'd like to call it ChannelValueObject? Isn't that a bit confusing and/or over-engineered?

Answer 1

Since both classes have the same getters and setters this can most easily be achieved using apache commons beanutils:

Channel channel = new Channel();
BeanUtils.copyProperties(channel, channelWrapper);

if you have more complex requirements (e.g. different property names or types), I would recommend dozer.