繁体   English   中英

如何在springboot中为Json构建java Pojo

[英]how to build java Pojo for Json in springboot

I have following Json and want to map this json as java pojo using spring webclient mono response. 在这个 json 中,键 data-1 和 en 不是常数,并且不断变化。 例如,它可以是 data-2 和 fr 或 data-3 和它。

{
    "data": {
        "attributes": {
            "data-1": {
                "en": [
                    {
                        "id": 1,
                        "code": "GI",
                        "order": 10
                        
                    },
                    {
                        "id": 2,
                        "code": "TH",
                        "order": 10
                    }
                ]
            }
        }
    }
}

DTO结构:

@JsonInclude(JsonInclude.Include.NON_NULL)
public class ListData {
    private Map<String, Attributes> data;
}

@JsonInclude(JsonInclude.Include.NON_NULL)
public class Attributes {
    private Map<String, Map<String, MyList>> attributes;
}

@JsonInclude(JsonInclude.Include.NON_NULL)
public class MyList {
    private List<Entry> entryList;
}

@JsonInclude(JsonInclude.Include.NON_NULL)
public class Entry {
    private String id;
    private String code;
    private String order;
    
}

这是反序列化 json 的 webclient。

return WebClient.create("test.com")
                .get()
                .uri(getTestUri())
                .header(HttpHeaders.ACCEPT, Constants.ACCEPT_TYPE)
                .retrieve()
                .bodyToMono(ListData.class)
                .block();

以下是错误:

org.springframework.core.codec.CodecException: Type definition error: [simple type, class com.test.ListAttributes]; nested exception is com.fasterxml.jackson.databind.exc.InvalidDefinitionException: Cannot construct instance of `com.test.ListAttributes` (no Creators, like default constructor, exist): cannot deserialize from Object value (no delegate- or property-based Creator)

有人可以帮我解决这个问题,如何解决。

{
    "data": {
        "attributes": {
            "data-1": {
                "en": [
                    {
                        "id": 1,
                        "code": "GI",
                        "order": 10
                        
                    },
                    {
                        "id": 2,
                        "code": "TH",
                        "order": 10
                    }
                ]
            }
        }
    }
}

“data-1”和“en”作为变量键意味着需要一个双重嵌套的Map 内部 map 将字符串映射到对象List (在您的特定情况下为Entry对象)。

因此,上述 JSON 的结构/类型映射到 Java 类型:

class Container {
  public Data data;
  public Container() {}
}

class Data {
  public Map<String, Map<String, List<Entry>>> attributes;
  public Data() {}
}

class Entry {
  public long id;
  public String code;
  public long order;
  public Enrty() {}
}

确保您的类型是默认可构造的,并且其字段可以由 Jackson 设置

暂无
暂无

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM