java – 使用@JsonSubTypes反序列化没有值 – 缺少属性错误
发布时间:2020-05-24 10:16:07 所属栏目:Java 来源:互联网
导读:我像这样反序列化jsons: { type:a, payload : {...}} 有效负载类型取决于类型.我的课: public class SthT extends Payload { @JsonProperty(type) private String type; @Valid private T payload; @
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我像这样反序列化jsons: {
"type":"a","payload" : {...}
}
有效负载类型取决于类型.我的课: public class Sth<T extends Payload> {
@JsonProperty("type")
private String type;
@Valid
private T payload;
@JsonTypeInfo(
use = JsonTypeInfo.Id.NAME,include = JsonTypeInfo.As.EXTERNAL_PROPERTY,property = "type",visible = true,defaultImpl = NoClass.class)
@JsonSubTypes({
@JsonSubTypes.Type(value = APayload.class,name = "a"),@JsonSubTypes.Type(value = BPayload.class,name = "b"),@JsonSubTypes.Type(value = CPayload.class,name = "c")})
public void setPayload(T payload) {
this.payload = payload;
}
public void setType(String type) {
this.type = type;
}
}
我也输入了没有有效载荷的“d”.如果我尝试反序列化: {
"type":"d","payload" : null
}
它可以工作,但它没有负载有效: {
"type":"d",}
如何让它与上一个例子一起工作? 我得到错误的Stacktrace: [error] Caused by: com.fasterxml.jackson.databind.JsonMappingException: Missing property 'payload' for external type id 'type [error] at [Source: N/A; line: -1,column: -1] [error] at com.fasterxml.jackson.databind.JsonMappingException.from(JsonMappingException.java:164) [error] at com.fasterxml.jackson.databind.DeserializationContext.mappingException(DeserializationContext.java:700) [error] at com.fasterxml.jackson.databind.deser.impl.ExternalTypeHandler.complete(ExternalTypeHandler.java:160) [error] at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeWithExternalTypeId(BeanDeserializer.java:690) [error] at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeWithExternalTypeId(BeanDeserializer.java:639) [error] at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:266) [error] at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:124) [error] at com.fasterxml.jackson.databind.ObjectMapper._readValue(ObjectMapper.java:2965) [error] at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:1587) [error] at com.fasterxml.jackson.databind.ObjectMapper.treeToValue(ObjectMapper.java:1931) [error] at play.libs.Json.fromJson(Json.java:47) 解决方法我也遇到过这个问题,并且使用Jackson提供的机制(自定义BeanDeserializer,BeanDeserializerModifier等)找不到优雅的解决方案.它看起来像处理外部类型ID的方式中的错误.我解决了以下问题: >将JSON tring反序列化为JsonNode; 我的代码如下所示: public <T> T decode(String json,Class<T> type) throws IOException {
JsonNode jsonNode = mapper.readTree(json);
if (jsonNode.isObject() && (jsonNode.get("payload") == null || jsonNode.get("payload").size() == 0)) {
ObjectNode objectNode = (ObjectNode) jsonNode;
objectNode.putNull("payload");
}
return mapper.treeToValue(jsonNode,type);
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