Question

现在我和杰克逊合作，我对此有一些疑问。

首先。我有两个服务，第一个是数据收集和发送服务，第二个接收这些数据，例如，将其记录到文件中。

所以，第一个服务有这样的类层次结构：

         +----ConcreteC
         |
Base ----+----ConcreteA
         |
         +----ConcreteB

第二个服务的类层次结构如下：

ConcreteAAdapter extends ConcreteA implements Adapter {}
ConcreteBAdapter extends ConcreteB implements Adapter {}
ConcreteCAdapter extends ConcreteC implements Adapter {}

第一项服务对ConcreteXAdapter一无所知。

我在第一项服务上发送数据的方式：

Collection<Base> data = new LinkedBlockingQueue<Base>()
JacksonUtils utils = new JacksonUtils();
data.add(new ConcreteA());
data.add(new ConcreteB());
data.add(new ConcreteC());
...
send(utils.marshall(data));
...

public class JacksonUtils {

    public byte[] marshall(Collection<Base> data) throws IOException {
        ByteArrayOutputStream out = new ByteArrayOutputStream() {
            @Override
            public byte[] toByteArray() {
                return buf;
            }
        };

        getObjectMapper().writeValue(out, data);
        return out.toByteArray();
    }
    protected ObjectMapper getObjectMapper() {
        return new ObjectMapper();
    }

    public Object unmarshall(byte[] json) throws IOException {
        return getObjectMapper().readValue(json, Object.class);
    }

    public <T> T unmarshall(InputStream source, TypeReference<T> typeReference) throws IOException {
        return getObjectMapper().readValue(source, typeReference);
    }

    public <T> T unmarshall(byte[] json, TypeReference<T> typeReference) throws IOException {
        return getObjectMapper().readValue(json, typeReference);
    }
}

所以，我想将json转换为ConcreteXAdapter的集合，而不是ConcreteX（ConcreteA -> ConcreteAAdapter, ConcreteB -> ConcreteBAdapter, ConcreteC -> ConcreteCAdapter）的集合。在我描述的情况下我想得到：

Collection [ConcreteAAdapter, ConcreteBAdapter, ConcreteCAdapter]

我该怎么做？

Answer 1

为此，您需要在JSON中传递其他信息：

@JsonTypeInfo(use=JsonTypeInfo.Id.NAME, 
      include=JsonTypeInfo.As.PROPERTY, property="@type")
class Base {
...
}

然后在序列化时，它将添加@type字段：

objectMapper.registerSubtypes(
            new NamedType(ConcreteAAdapter.class, "ConcreteA"),
            new NamedType(ConcreteBAdapter.class, "ConcreteB"),
            new NamedType(ConcreteCAdapter.class, "ConcreteC")
            );

// note, that for lists you need to pass TypeReference explicitly
objectMapper.writerWithType(new TypeReference<List<Base>>() {})
     .writeValueAsString(someList);


    {
      "@type" : "ConcreteA",
      ...
    }

反序列化将是：

    objectMapper.registerSubtypes(
            new NamedType(ConcreteA.class, "ConcreteA"),
            new NamedType(ConcreteB.class, "ConcreteB"),
            new NamedType(ConcreteC.class, "ConcreteC")
            );
    objectMapper.readValue(....)

更多信息：http://wiki.fasterxml.com/JacksonPolymorphicDeserialization

Answer 2

我是如何解决这个问题的。这是一个示例项目的类图： class diagram

所以我希望在反序列化后获得ConcreteAAdapter表单ConcreteA。

我的解决方案是扩展ClassNameIdResolver以添加将基类对象反序列化为子类型对象的功能（子类型不添加额外的功能和附加字段）。

以下是为反序列化创建ObjectMapper的代码：

protected ObjectMapper getObjectMapperForDeserialization() {
        ObjectMapper mapper = new ObjectMapper();

        StdTypeResolverBuilder typeResolverBuilder = new ObjectMapper.DefaultTypeResolverBuilder(ObjectMapper.DefaultTyping.OBJECT_AND_NON_CONCRETE);
        typeResolverBuilder = typeResolverBuilder.inclusion(JsonTypeInfo.As.PROPERTY);
        typeResolverBuilder.init(JsonTypeInfo.Id.CLASS, new ClassNameIdResolver(SimpleType.construct(Base.class), TypeFactory.defaultInstance()) {
            private HashMap<Class, Class> classes = new HashMap<Class, Class>() {
                {
                    put(ConcreteA.class, ConcreteAAdapter.class);
                    put(ConcreteB.class, ConcreteBAdapter.class);
                    put(ConcreteC.class, ConcreteCAdapter.class);
                }
            };

            @Override
            public String idFromValue(Object value) {
                return (classes.containsKey(value.getClass())) ? value.getClass().getName() : null;
            }

            @Override
            public JavaType typeFromId(String id) {
                try {
                    return classes.get(Class.forName(id)) == null ? super.typeFromId(id) : _typeFactory.constructSpecializedType(_baseType, classes.get(Class.forName(id)));
                } catch (ClassNotFoundException e) {
                    // todo catch the e
                }
                return super.typeFromId(id);
            }
        });
        mapper.setDefaultTyping(typeResolverBuilder);
        return mapper;
    }

这是一个为序列化创建ObjectMapper的代码：

protected ObjectMapper getObjectMapperForSerialization() {
    ObjectMapper mapper = new ObjectMapper();

    StdTypeResolverBuilder typeResolverBuilder = new ObjectMapper.DefaultTypeResolverBuilder(ObjectMapper.DefaultTyping.OBJECT_AND_NON_CONCRETE);
    typeResolverBuilder = typeResolverBuilder.inclusion(JsonTypeInfo.As.PROPERTY);
    typeResolverBuilder.init(JsonTypeInfo.Id.CLASS, new ClassNameIdResolver(SimpleType.construct(Base.class), TypeFactory.defaultInstance()));
    mapper.setDefaultTyping(typeResolverBuilder);

    return mapper;
}

测试代码：

public static void main(String[] args) throws IOException {
    JacksonUtils JacksonUtils = new JacksonUtilsImpl();

    Collection<Base> data = new LinkedBlockingQueue<Base>();
    data.add(new ConcreteA());
    data.add(new ConcreteB());
    data.add(new ConcreteC());

    String json = JacksonUtils.marshallIntoString(data);

    System.out.println(json);

    Collection<? extends Adapter> adapters = JacksonUtils.unmarshall(json, new TypeReference<ArrayList<Adapter>>() {});

    for (Adapter adapter : adapters) {
        System.out.println(adapter.getClass().getName());
    }
}

JacksonUtils类的完整代码：

public class JacksonUtilsImpl implements JacksonUtils {

    @Override
    public byte[] marshall(Collection<Base> data) throws IOException {
        ByteArrayOutputStream out = new ByteArrayOutputStream() {
            @Override
            public byte[] toByteArray() {
                return buf;
            }
        };

        getObjectMapperForSerialization().writerWithType(new TypeReference<Collection<Base>>() {}).writeValue(out, data);
        return out.toByteArray();
    }

    @Override
    public String marshallIntoString(Collection<Base> data) throws IOException {
        return getObjectMapperForSerialization().writeValueAsString(data);
    }

    protected ObjectMapper getObjectMapperForSerialization() {
        ObjectMapper mapper = new ObjectMapper();

        StdTypeResolverBuilder typeResolverBuilder = new ObjectMapper.DefaultTypeResolverBuilder(ObjectMapper.DefaultTyping.OBJECT_AND_NON_CONCRETE);
        typeResolverBuilder = typeResolverBuilder.inclusion(JsonTypeInfo.As.PROPERTY);
        typeResolverBuilder.init(JsonTypeInfo.Id.CLASS, new ClassNameIdResolver(SimpleType.construct(Base.class), TypeFactory.defaultInstance()));
        mapper.setDefaultTyping(typeResolverBuilder);

        return mapper;
    }

    protected ObjectMapper getObjectMapperForDeserialization() {
        ObjectMapper mapper = new ObjectMapper();

        StdTypeResolverBuilder typeResolverBuilder = new ObjectMapper.DefaultTypeResolverBuilder(ObjectMapper.DefaultTyping.OBJECT_AND_NON_CONCRETE);
        typeResolverBuilder = typeResolverBuilder.inclusion(JsonTypeInfo.As.PROPERTY);
        typeResolverBuilder.init(JsonTypeInfo.Id.CLASS, new ClassNameIdResolver(SimpleType.construct(Base.class), TypeFactory.defaultInstance()) {
            private HashMap<Class, Class> classes = new HashMap<Class, Class>() {
                {
                    put(ConcreteA.class, ConcreteAAdapter.class);
                    put(ConcreteB.class, ConcreteBAdapter.class);
                    put(ConcreteC.class, ConcreteCAdapter.class);
                }
            };

            @Override
            public String idFromValue(Object value) {
                return (classes.containsKey(value.getClass())) ? value.getClass().getName() : null;
            }

            @Override
            public JavaType typeFromId(String id) {
                try {
                    return classes.get(Class.forName(id)) == null ? super.typeFromId(id) : _typeFactory.constructSpecializedType(_baseType, classes.get(Class.forName(id)));
                } catch (ClassNotFoundException e) {
                    // todo catch the e
                }
                return super.typeFromId(id);
            }
        });
        mapper.setDefaultTyping(typeResolverBuilder);
        return mapper;
    }

    @Override
    public Object unmarshall(byte[] json) throws IOException {
        return getObjectMapperForDeserialization().readValue(json, Object.class);
    }

    @Override
    public <T> T unmarshall(InputStream source, TypeReference<T> typeReference) throws IOException {
        return getObjectMapperForDeserialization().readValue(source, typeReference);
    }

    @Override
    public <T> T unmarshall(byte[] json, TypeReference<T> typeReference) throws IOException {
        return getObjectMapperForDeserialization().readValue(json, typeReference);
    }

    @Override
    public <T> Collection<? extends T> unmarshall(String json, Class<? extends Collection<? extends T>> klass) throws IOException {
        return getObjectMapperForDeserialization().readValue(json, klass);
    }


    @Override
    public <T> Collection<? extends T> unmarshall(String json, TypeReference typeReference) throws IOException {
        return getObjectMapperForDeserialization().readValue(json, typeReference);
    }
}

Answer 3

我发现programmerbruce的方法是最清晰，最容易上手的（下面的例子）。我从他对相关问题的回答中得到了相关信息： https://stackoverflow.com/a/6339600/1148030 和相关的博客文章： http://programmerbruce.blogspot.fi/2011/05/deserialize-json-with-jackson-into.html

另请查看这个友好的维基页面（也在Eugene Retunsky的回答中提到）： https://github.com/FasterXML/jackson-docs/wiki/JacksonPolymorphicDeserialization

另一个不错的维基页面：https://github.com/FasterXML/jackson-docs/wiki/JacksonMixInAnnotations

这是一个简短的例子，可以给你一个想法：

像这样配置ObjectMapper：

    mapper.getDeserializationConfig().addMixInAnnotations(Base.class, BaseMixin.class);
    mapper.getSerializationConfig().addMixInAnnotations(Base.class, BaseMixin.class);

示例BaseMixin类（易于定义为内部类。）

@JsonTypeInfo(use=JsonTypeInfo.Id.NAME, include=JsonTypeInfo.As.PROPERTY, property="type")
@JsonSubTypes({
    @JsonSubTypes.Type(value=ConcreteA.class, name="ConcreteA"),
    @JsonSubTypes.Type(value=ConcreteB.class, name="ConcreteB")
})  
private static class BaseMixin {
}

在第二项服务上，您可以像这样定义BaseMixin：

@JsonTypeInfo(use=JsonTypeInfo.Id.NAME, include=JsonTypeInfo.As.PROPERTY, property="type")
@JsonSubTypes({
    @JsonSubTypes.Type(value=ConcreteAAdapter.class, name="ConcreteA"),
    @JsonSubTypes.Type(value=ConcreteBAdapter.class, name="ConcreteB")
})  
private static class BaseMixin {
}

使用Jackson将Json反序列化为另一个类层次结构

3 个答案: