【问题标题】:Deserializing a Java Object that extends an ArrayList with Jackson使用 Jackson 反序列化扩展 ArrayList 的 Java 对象
【发布时间】:2014-09-30 20:39:47
【问题描述】:

我有以下类,我可以以我想要的方式序列化。

@JsonSerialize(using = MySerializer.class)
@JsonDeserialize(using = MyDeserialize.class)
public class TryToSerialize extends ArrayList<String>
{
    private int number;
    private String word;
    public TryToSerialize(){}
    public TryToSerialize(int number, String word){
        this.number = number;
        this.word = word;
    }

    @JsonProperty
    public int getNumber() {
        return number;
    }

    public void setNumber(int number) {
        this.number = number;
    }
    @JsonProperty
    public String getWord() {
        return word;
    }

    public void setWord(String word) {
        this.word = word;
    }
}

已经用下面的序列化器序列化了

class MySerializer extends JsonSerializer<TryToSerialize> 
{
    @Override
    public void serialize(TryToSerialize toSerialize, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException 
    {
       jgen.writeStartObject();
        jgen.writeNumberField("number", toSerialize.getNumber());
        jgen.writeStringField("word", toSerialize.getWord());
        provider.defaultSerializeField("values", toSerialize.iterator(), jgen);
        jgen.writeEndObject();
    }
}

此代码会将对象序列化为以下内容:

{"number":10,"word":"Working","values":["first","second","third"]}

现在我正在尝试反序列化它,但我不确定如何处理 ArrayList 部分的处理。到目前为止,我有这样的事情:

class MyDeserialize extends JsonDeserializer<TryToSerialize> {
    @Override
    public TryToSerialize deserialize(JsonParser jp, DeserializationContext ctxt) 
      throws IOException, JsonProcessingException {
        JsonNode node = jp.getCodec().readTree(jp);

        int number = (Integer) ((IntNode) node.get("number")).numberValue();
        String word = node.get("word").asText();


        return new TryToSerialize(number, word);
    }
}

任何帮助将不胜感激。

【问题讨论】:

标签: java json inheritance jackson unmarshalling


【解决方案1】:

到目前为止,您拥有的一个选项是,因为您有可用的 JsonNode,并且您知道您已将列表序列化为属性 values,以迭代 values 属性并将内容添加到您的类.

class MyDeserializer extends JsonDeserializer<TryToSerialize> {
    @Override
    public TryToSerialize deserialize(JsonParser jp, DeserializationContext ctxt)
            throws IOException, JsonProcessingException {
        JsonNode node = jp.getCodec().readTree(jp);

        int number = (Integer) ((IntNode) node.get("number")).getNumberValue();
        String word = node.get("word").asText();

        TryToSerialize deserialized = new TryToSerialize(number, word);
        JsonNode valuesNode = node.get("values");
        if (valuesNode.isArray()) {
            Iterator<JsonNode> arrayIterator = ((ArrayNode) valuesNode).getElements();
            while (arrayIterator.hasNext()) {
                deserialized.add(arrayIterator.next().getTextValue());
            }

        }

        return deserialized;
    }
}

【讨论】:

  • 正是我想要的。
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