java – 使用@JsonSubTypes反序列化没有值 – 缺少属性错误

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;
>如果不存在required属性,则手动插入空节点;
>将JsonNode映射到我想要的值类型.

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);
}