如何在构建器中获取超类字段?
父母:
type GetUser = (dbUser: DbUser | null) => GqlUser | null;
const getUser: GetUser = (dbUser) => {
if(!dbUser) return null;
return ... //gql user
};
const dbUser = { ... some fields} //not null for sure
getUser(dbUser) //here I want typescript to know that the result will not be null
孩子:
package com.goomo.sso.dto;
import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor;
import java.io.Serializable;
@AllArgsConstructor
@NoArgsConstructor
@Data
public class BaseResponseModel implements Serializable {
private static final long serialVersionUID = 1L;
private String status;
private StatusMessage statusMessage;
}
我无法在构建器中获取父类的字段。您能建议我实现该目标的正确方法吗?