Java XML Jackson解析XML并忽略某些元素

Question

我想将model_tree的字段变成对象。

    <form>
      <model_trees>
        <entry>
            <tree_kind>kind</tree_kind>
            <age>10</age>
        </entry>
      </model_trees>
    </form>

    @JsonIgnoreProperties(ignoreUnknown=true)
    @JacksonXmlRootElement(localName = "model_trees")
    public class ModelTrees extends BaseEntity {
      @JacksonXmlProperty(localName = "tree_kind")
      private String treeKind;
      @JacksonXmlProperty
      private int age;
    }

有没有办法不为“表单”/“输入”元素创建其他类别？

Answer 1

尝试自定义求职者，例如

class DeSerializer extends StdDeserializer<ModelTrees> {

    protected DeSerializer() {
        super(ModelTrees.class);
    }

    @Override
    public ModelTrees deserialize(JsonParser jp, DeserializationContext ctxt) throws IOException, JsonProcessingException {
        JsonNode node = jp.getCodec().readTree(jp);

        JsonNode treesNode = node.get("model_trees");

        JsonNode entryNode = treesNode.get("entry");

        return new ModelTrees(entryNode.get("tree_kind").textValue(), entryNode.get("age").asInt());
    }
}

模型类：

public class ModelTrees {
    private String treeKind;
    private int age;

    public ModelTrees(String treeKind, int age) {
        this.treeKind = treeKind;
        this.age = age;
    }
}

注册自定义映射器：

XmlMapper mapper = new XmlMapper();
SimpleModule module = new SimpleModule("configModule",   com.fasterxml.jackson.core.Version.unknownVersion());
module.addDeserializer(ModelTrees.class, new DeSerializer());
mapper.registerModule(module);