java - 使用jackson xml映射器将xml反序列化为pojo

Question

我正在使用 Jackson XML 映射器将 XML 反序列化为 POJO。XML看起来像

<person>
 <agency>
        <phone>111-111-1111</phone>
 </agency>
</person>

我的课看起来像

class Person
{
 @JacksonXmlProperty(localName="agency", namespace="namespace")
 private Agency agency;
 //getter and setter
}
class Agency
{
 @JacksonXmlElementWrapper(useWrapping = false)
 @JacksonXmlProperty(localName="phone", namespace="namespace")
 private List<AgencyPhone> phones;
 //getter and setter
}
class AgencyPhone
{
  private Phone phone;
  //getter and setter
}
class Phone
{
 private String number;
 //getter and setter
}

我想将电话号码设置为电话类中的号码。我无法更改 XML 或类的结构方式。我收到无法构造resolved.agency.AgencyPhone错误实例，我创建了一个 AgencyPhone 构造函数

class AgencyPhone{
{
  private Phone phone;
  public AgencyPhone(Phone phone)
  {
      this.phone = phone;
   }
  }

但这没有用。那么如何反序列化为嵌套实例。

Answer 1

您可以编写自己的自定义反序列化器来实现此目的。以下是帮助您入门的代码：

import com.fasterxml.jackson.core.JsonParseException;
import com.fasterxml.jackson.core.JsonParser;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.DeserializationContext;
import com.fasterxml.jackson.databind.JsonMappingException;
import com.fasterxml.jackson.databind.deser.std.StdDeserializer;
import com.fasterxml.jackson.databind.module.SimpleModule;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;
import java.io.IOException;

public class Test {
  public static void main(String[] args) throws JsonParseException, JsonMappingException, IOException {
    XmlMapper mapper = new XmlMapper();
    final SimpleModule module = new SimpleModule("configModule",   com.fasterxml.jackson.core.Version.unknownVersion());
    module.addDeserializer(Person.class, new DeSerializer());
    mapper.registerModule(module);
    // Person readValue = mapper.readValue(<xml source>);
  }
}

class DeSerializer extends StdDeserializer<Person> {

  protected DeSerializer() {
    super(Person.class);
  }

  @Override
  public Person deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
    // use p.getText() and p.nextToken to navigate through the xml and construct Person object
    return new Person();

  }
}