在Java中将XML转换为JSON的最快方法

   <groupId>com.fasterxml.jackson.dataformat</groupId>
   <artifactId>jackson-dataformat-xml</artifactId>
   <version>2.13.0</version>
</dependency>

import com.fasterxml.jackson.databind.JsonNode;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;

public class XmlToJsonConverter {
    public static void main(String[] args) throws Exception {
        String xmlString = "<root><name>John</name><age>30</age></root>";

        ObjectMapper xmlMapper = new XmlMapper();
        JsonNode jsonNode = xmlMapper.readTree(xmlString.getBytes());

        ObjectMapper jsonMapper = new ObjectMapper();
        String jsonString = jsonMapper.writeValueAsString(jsonNode);

        System.out.println(jsonString);
    }
}

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