使用Jackson将XML转换为具有不同属性名称的JSON

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>2.12.5</version>
</dependency>

public class Person {
    @JsonProperty("name")
    private String fullName;

    @JsonProperty("age")
    private int years;

    // Getters and setters
}

XmlMapper xmlMapper = new XmlMapper();
String xml = "<person><name>John Doe</name><age>30</age></person>";

Person person = xmlMapper.readValue(xml, Person.class);
ObjectMapper jsonMapper = new ObjectMapper();
String json = jsonMapper.writeValueAsString(person);

System.out.println(json);

{
  "name": "John Doe",
  "age": 30
}