是否可以让杰克逊将嵌套对象序列化为字符串

Question

鉴于这些类：

@Value
private static class Message {
    private final String type;
    private final MyType message;
}

@Value
public class MyType {
    private final String foo;
}

杰克逊将产生：

{
  "Type" : "Test",
  "Message" : {"foo" : "bar"}
}

我可以给 Jackson 提供某种类型的注释或指令，要求它将嵌套的复杂类型序列化为字符串，例如所需的 JSON 将是：

{
  "Type" : "Test",
  "Message" : "{\"foo\" : \"bar\"}"
}

我在消息字段上尝试了这两个注释：

 @JsonFormat(shape = JsonFormat.Shape.STRING)
 @JsonSerialize(as=String.class)

两者都没有预期的影响。现在我的“hack”是在施工时做到这一点：

return new Message("Test", mapper.writeValueAsString(new MyType("bar")));

我想我可以编写一个自定义序列化程序，但我想知道这是否是某种内置的标准行为。我的用例是我正在构建一个 JSON 有效负载，它应该有一个字符串消息包含在它本身包含JSON。

环境

Jackson 版本是 2.9.0，在 Java 10 上使用 Spring Boot 2。

Answer 1

可以使用自定义序列化器来完成：

class EscapedJsonSerializer extends StdSerializer<Object> {
    public EscapedJsonSerializer() {
        super((Class<Object>) null);
    }


    @Override
    public void serialize(Object value, JsonGenerator gen, SerializerProvider provider) throws IOException {
        StringWriter str = new StringWriter();
        JsonGenerator tempGen = new JsonFactory().setCodec(gen.getCodec()).createGenerator(str);
        if (value instanceof Collection || value.getClass().isArray()) {
            tempGen.writeStartArray();
            if (value instanceof Collection) {
                for (Object it : (Collection) value) {
                    writeTree(gen, it, tempGen);
                }
            } else if (value.getClass().isArray()) {
                for (Object it : (Object[]) value) {
                    writeTree(gen, it, tempGen);
                }
            }
            tempGen.writeEndArray();
        } else {
            provider.defaultSerializeValue(value, tempGen);
        }
        tempGen.flush();
        gen.writeString(str.toString());
    }


    @Override
    public void serializeWithType(Object value, JsonGenerator gen, SerializerProvider serializers, TypeSerializer typeSer) throws IOException {
        StringWriter str = new StringWriter();
        JsonGenerator tempGen = new JsonFactory().setCodec(gen.getCodec()).createGenerator(str);
        writeTree(gen, value, tempGen);
        tempGen.flush();
        gen.writeString(str.toString());
    }

    private void writeTree(JsonGenerator gen, Object it, JsonGenerator tempGen) throws IOException {
        ObjectNode tree = ((ObjectMapper) gen.getCodec()).valueToTree(it);
        tree.set("@class", new TextNode(it.getClass().getName()));
        tempGen.writeTree(tree);
    }
}

和反序列化器：

class EscapedJsonDeserializer extends JsonDeserializer<Object> implements ContextualDeserializer {
    private final Map<JavaType, JsonDeserializer<Object>> cachedDeserializers = new HashMap<>();

    @Override
    public Object deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
        throw new IllegalArgumentException("EscapedJsonDeserializer should delegate deserialization for concrete class");

    }

    @Override
    public JsonDeserializer<?> createContextual(DeserializationContext ctxt, BeanProperty property) throws JsonMappingException {
        JavaType type = (ctxt.getContextualType() != null) ?
                ctxt.getContextualType() : property.getMember().getType();
        return cachedDeserializers.computeIfAbsent(type, (a) -> new InnerDeserializer(type));
    }

    private class InnerDeserializer extends JsonDeserializer<Object> {
        private final JavaType javaType;

        private InnerDeserializer(JavaType javaType) {
            this.javaType = javaType;
        }

        @Override
        public Object deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
            String string = p.readValueAs(String.class);
            return ((ObjectMapper) p.getCodec()).readValue(string, javaType);
        }

        @Override
        public Object deserializeWithType(JsonParser p, DeserializationContext ctxt, TypeDeserializer typeDeserializer)
                throws IOException {

            String str = p.readValueAs(String.class);


            TreeNode root = ((ObjectMapper) p.getCodec()).readTree(str);
            Class clz;
            try {
                clz = Class.forName(((TextNode) root.get("@class")).asText());
                Object newJsonNode = p.getCodec().treeToValue(root, clz);
                return newJsonNode;
            } catch (ClassNotFoundException e) {
                throw new RuntimeException(e);
            }
        }
    }
}

该字段应使用@JsonSerialize 和@JsonDeserialize 注释（如果需要）

class Outer {
    @JsonTypeInfo(include = JsonTypeInfo.As.PROPERTY, use = JsonTypeInfo.Id.CLASS)
    @JsonSerialize(using = EscapedJsonSerializer.class)
    @JsonDeserialize(using = EscapedJsonDeserializer.class)
    public Foo val;
}

它适用于简单的集合（列表、数组），并且在某种程度上适用于多态性，尽管对于特定的多态性相关问题可能需要更精细的解决方案。 示例输出如下所示：

{"val":"{\"foo\":\"foo\",\"@class\":\"org.test.Foo\"}"}
{"val":"{\"foo\":\"foo\",\"bar\":\"bar\",\"@class\":\"org.test.Bar\"}"}

Answer 2

我也找不到内置解决方案，最终编写了自定义转换器：

public class ObjectToJsonStringConverter extends StdConverter<Object, String> {

    private final ObjectMapper objectMapper = new ObjectMapper();

    @Override
    public String convert(Object value) {
        try {
            return objectMapper.writeValueAsString(value);
        } catch (JsonProcessingException e) {
            throw new IllegalStateException(e);
        }
    }
}

用法：

@Value
private static class Message {
    private final String type;

    @JsonSerialize(converter = ObjectToJsonStringConverter.class)
    private final MyType message;
}