如何在将Xml文件转换为json文件时忽略某些字符

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

import java.io.File;
import java.io.IOException;

public class XmlToJsonConverter {
    public static void main(String[] args) {
        try {
            // 创建XmlMapper对象
            XmlMapper xmlMapper = new XmlMapper();
            
            // 忽略特定字符的处理逻辑
            xmlMapper.enable(SerializationFeature.INDENT_OUTPUT);
            xmlMapper.setDefaultUseWrapper(false);
            
            // 读取XML文件并转换为JSON
            File xmlFile = new File("input.xml");
            Object xmlObject = xmlMapper.readValue(xmlFile, Object.class);
            String jsonString = xmlMapper.writeValueAsString(xmlObject);
            
            // 输出JSON结果
            System.out.println(jsonString);
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}