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Java 显示所需的json dto类_Java_Json_Spring_Dto - Fatal编程技术网

Java 显示所需的json dto类

Java 显示所需的json dto类,java,json,spring,dto,Java,Json,Spring,Dto,创建了一个dto类,通过服务jsonschema2pojo.org/处理json。 结果是一个包含其他类的类,例如: public class GeneralExmp{ private Strig name; // GetAndSet public class Exmp2{ private Strig id; private Map<String, Object> additionalProperties = new HashMap<String, Object&g

创建了一个dto类,通过服务jsonschema2pojo.org/处理json。 结果是一个包含其他类的类,例如:

public  class GeneralExmp{
 private Strig name;
 // GetAndSet
  public class Exmp2{
 private Strig id;
private Map<String, Object> additionalProperties = new HashMap<String, Object>();
// GetAndSet

}
 public class Exmp3{
 private Strig val;
private Map<String, Object> additionalProperties = new HashMap<String, Object>();
// GetAndSet

}
}
json获得的类:

@Service
public class InstrumentGetService {
    RestTemplate restTemlate;
        
        public List<Params> getInfo(String token,
                 String order) {
            final var url = String.format(
                    "https://example.ru/rest/getOrderStatus.do?token=%s&orderId=%s", token, order);
            GeneralExmp dto = restTemlate.getForObject(url, GeneralExmp.class);
            return Collections.singletonList(toModel(dto));
        }
告诉我如何正确实现toModel方法?我无法将get和set从Exmp类获取到Params构造函数中

@Service
public class InstrumentGetService {
    RestTemplate restTemlate;
        
        public List<Params> getInfo(String token,
                 String order) {
            final var url = String.format(
                    "https://example.ru/rest/getOrderStatus.do?token=%s&orderId=%s", token, order);
            GeneralExmp dto = restTemlate.getForObject(url, GeneralExmp.class);
            return Collections.singletonList(toModel(dto));
        }
public Params toModel(GeneralExmp dto) {
            //??
            
            return new Params(//? , //?);
        }