使用 Jackson XML 映射器将 Java List 序列化为 XML

java xml jackson pojo xmlmapper

2022-09-01 16:37:13

嗨，我需要使用Jackson-dataformat XMLMapper从JAVA创建一个XML。XML 应该像

<Customer>
  <id>1</id>
  <name>Mighty Pulpo</name>
    <addresses>
      <city>austin</city>
      <state>TX</state>
    </addresses>
    <addresses>
      <city>Hong Kong</city>
      <state>Hong Kong</state>
    </addresses>
</Customer>

但我总是像一个额外的“<地址> </addresses>”标签一样。

<Customer>
  <id>1</id>
  <name>Mighty Pulpo</name>
<addresses>
    <addresses>
      <city>austin</city>
      <state>TX</state>
    </addresses>
    <addresses>
      <city>Hong Kong</city>
      <state>Hong Kong</state>
    </addresses>
<addresses>
</Customer>

我正在使用下面的代码来创建XML

JaxbAnnotationModule jaxbAnnotationModule = new JaxbAnnotationModule();
XmlMapper mapper = new XmlMapper();
mapper.enable(SerializationFeature.INDENT_OUTPUT);
mapper.registerModule(jaxbAnnotationModule);
mapper.registerModule(new GuavaModule());
String xml = mapper.writeValueAsString(customer);
System.out.println(xml);

请问有人可以帮我吗？我怎么能删除额外的标签请.我试图使用@XmlElement，但它没有帮助。蒂亚！！

答案 1

请尝试以下代码

@JacksonXmlRootElement(localName = "customer") 
class Customer {

    @JacksonXmlProperty(localName = "id")
    private int id;
    @JacksonXmlProperty(localName = "name")
    private String  name;

    @JacksonXmlProperty(localName = "addresses")
    @JacksonXmlElementWrapper(useWrapping = false)
    private Address[] address;

    //getters, setters, toString
}

class Address {

    @JacksonXmlProperty(localName = "city")
    private String city;

    @JacksonXmlProperty(localName = "state")
    private String state;
    // getter/setter 
}

答案 2

此设置更改默认包装行为，如果您不想处理代码中所有位置的批注。

XmlMapper mapper = new XmlMapper();
mapper.setDefaultUseWrapper(false);