如何只返回JSON对象中的某些字段？

Question

I currently have a function that returns a JSON version of an object:

public class Debate extends Controller
{
    public static Result viewArgument(Long id)
    {
        ...
        return ok(Json.toJson(Argument.get(id)));
    }
}

This Argument object has some confidential information that should not be exposed to the client, however. How can I select only the fields id and summary to be returned in the resulting JSON object?

Answer 1

You can copy the Argument 's id and summary into a DTO (Data Transfer Object). Then turn that into JSON to be sent over the wire.

public class ArgumentDto {
    public Long id;
    public String summary;
}

public class Debate extends Controller
{
    public static Result viewArgument(Long id)
    {
        ...
        Argument originalArgument = Argument.get(id);
        ArgumentDto argument = new ArgumentDto();
        dto.id = originalArgument.id;
        dto.summary = originalArgument.summary;
        return ok(Json.toJson(dto));
    }
}

Answer 2

有一个更简单的答案，只需将@JsonIgnore添加到其他字段。