使用@JsonSubTypes 反序列化无值 - 缺少属性错误

问题描述

我像这样反序列化json:

I deserialize jsons like this:

{
  "type":"a",
  "payload" : {...}
}

其中有效载荷类型取决于类型.我的班级:

where payload type depends on type. My class:

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;
    }

}

我还输入了没有有效载荷的d".如果我尝试反序列化:

I have also type "d" with no payload. If I try to deserialize:

{
  "type":"d",
  "payload" : null
}

它可以工作，但在没有有效负载的情况下无法工作:

it works but it doesn't work with no payload:

{
  "type":"d",
}

如何让它与上一个例子一起工作?

How to make it work with last example?

我得到的错误堆栈跟踪:

Stacktrace of error that I get:

[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)

推荐答案

我也遇到过这个问题，使用 Jackson 提供的机制(自定义BeanDeserializer，BeanDeserializerModifier 等).

I've also encountered this issue, and could not find an elegant solution using the mechanisms provided by Jackson (custom BeanDeserializer, BeanDeserializerModifier, etc.).

这看起来像是处理外部类型 ID 方式的错误.我通过以下方式解决了这个问题:

It looks like a bug in the way external type IDs are handled. I worked around it by:

1. 将 JSON 字符串反序列化为 JsonNode;
2. 如果所需的属性不存在，则手动插入一个 null 节点；
3. 将 JsonNode 映射到我想要的值类型.

1. deserializing the JSON tring to a JsonNode;
2. manually inserting a null node if the required property is not present;
3. mapping the JsonNode to my desired value type.

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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