Question

假设我有以下结构：

public class SomeClass {
private String id;
@JsonProperty("key-value")
private Map<String, Object> keyValue;}

显然，它将被序列化为

 {
   "id" : "id1", 
   "key-value" : 
    {"key1" : "value1",
    "key2 : "value2"}
 }

是否有可能像这样代表它？

 {
   "id" : "id1", 
   "key1" : "value1",
   "key2 : "value2"
 }

提前致谢！

Answer 1

在Jackson的自定义序列化器的帮助下很有可能：

将@JsonSerialize annoation添加到您的POJO：

（还添加了必要的ctor和getters）

@JsonSerialize(using = SomeClassSerializer.class)
public static class SomeClass {
    private String id;
    @JsonProperty("key-value")
    private Map<String, Object> keyValue;

    public SomeClass(String id, Map<String, Object> keyValue) {
        this.id = id;
        this.keyValue = keyValue;
    }
    public String getId() { return id; }
    public Map<String, Object> getKeyValue() { return keyValue; }
}

自定义序列化程序如下所示：

public class SomeClassSerializer extends JsonSerializer<SomeClass>
{
    @Override
    public void serialize(SomeClass sc, JsonGenerator gen, SerializerProvider serializers)
            throws IOException, JsonProcessingException
    {
        gen.writeStartObject();
        // write id propertry
        gen.writeStringField("id", sc.getId());   
        // loop on keyValue entries, write each as property
        for (Map.Entry<String, Object> keyValueEntry : sc.getKeyValue().entrySet()) {
            gen.writeObjectField(keyValueEntry.getKey(), keyValueEntry.getValue());
        }
        gen.writeEndObject();
    }
}

调用杰克逊的映射器是以通常的方式完成的：

public static void main(String[] args)
{
    Map<String, Object> keyValue = new HashMap<>();
    keyValue.put("key1", "value1");
    keyValue.put("key2", "value2");
    keyValue.put("key3", new Integer(10));
    SomeClass sc = new SomeClass("id1", keyValue);

    try {
        new ObjectMapper().writeValue(System.out, sc);
    } catch (IOException e) {
        e.printStackTrace();
    }
}

输出：

{"id":"id1","key1":"value1","key2":"value2","key3":10}

杰森杰克逊不寻常的地图序列化

1 个答案: