如何过滤JAX-RS服务的响应？

Question

I got the following value object:

@XmlRootElement
public class Movie{
  public String name;
  public Date releaseDate;
  public List<Actors> actors;
}

and i got the following service

@GET
@Produces(MediaType.APPLICATION_JSON)
public List<Movie> moviesByYear(int year){
//return all movies by year
}

The movies are searched in the database by some ORM framework. My question is: I want to filter the response, to not return the actors list (because this field is not relevant, and makes the response larger). Of course I can

for(Movie movie: movies){
  movie.actors = null;
}

but this will escalate quickly if I want to remove multiple fields.

Answer 1

If you never want to include the actors field in your response, you can annotate the field with @XmlTransient . See the JavaDoc for more details.

Otherwise, you could wrap the Movie object into a wrapper object that doesn't expose the actors field.

Answer 2

如果服务没有执行此操作的方法，则必须更改服务代码本身。