Jackson XML Serialization XMLElementRef名称

Question

您好我需要将fastxml.Jackson Serializer配置为与默认的JAXB Serializer完全兼容。

我有两个类的以下案例：

@XmlRootElement(name = "elA")
public class A {
    @XmlElementRef
    public Collection<B> getBs() {
        return this.bs;
    }
}

@XmlRootElement(name = "elB")
public class B {

}

My Object Mapper配置如下所示：

XmlMapper objectMapper = new XmlMapper();
JaxbAnnotationModule jaxbAnnotationModule = new JaxbAnnotationModule();
objectMapper.registerModule(jaxbAnnotationModule);

objectMapper.setDefaultUseWrapper(false);
objectMapper.configure(SerializationFeature.INDENT_OUTPUT,true);
objectMapper.configure(SerializationFeature.WRAP_ROOT_VALUE,false);
objectMapper.configure(SerializationFeature.CLOSE_CLOSEABLE, true);
objectMapper.setSerializationInclusion(Include.NON_ABSENT);

JaxbAnnotationIntrospector jaxbAnnotationIntrospector = new XmlJaxbAnnotationIntrospector(objectMapper.getTypeFactory());
objectMapper.setAnnotationIntrospector(jaxbAnnotationIntrospector);

我需要为Jackson Serializer配置ObjectMapper以生成以下XML

<elA>
   <elB></elB>
   <elB></elB>
</elA>

但我明白了：

<A>
    <bs></bs>
    <bs></bs>
</A>

如何配置Object Mapper以获得预期结果？谢谢。

我不能改变类和注释，因为这是框架的一部分。

我唯一可以影响的是ObjectMapper和Jackson配置。

Answer 1

@XmlRootElement(name = "elA")
public class ElA{
@XmlElementRef
public Collection<B> getElB() {
    return this.elb;
}
}

@XmlRootElement(name = "elB")
public class ElB{

}