由于jackson,Java二进制序列化失败

时间:2015-05-19 15:34:33

标签: java json serialization jackson

我使用jackson 2将json转换为java对象。到现在为止还挺好。但我也使用hazelcast来分发集群中的对象。因此,所有bean都必须是java.io.Serializable。当我从json中读取Object时:

ObjectMapper mapper = new ObjectMapper();
mapper.addMixInAnnotations(AbstractBean.class, MongoIdMixIn.class);

// this is to prevent from failing on missing type class property: @JsonProperty("@class")
Object tgtObject = targetClass.newInstance();
mapper.readerForUpdating(tgtObject).readValue(dbo.toString());

// put into hazelcast map
target.put(dbo.get(keyColumn), tgtObject); 

我将从hazelcast获得例外:

java.io.NotSerializableException: com.fasterxml.jackson.databind.deser.impl.TypeWrappedDeserializer

我想知道com.fasterxml.jackson.databind.deser.impl.TypeWrappedDeserializer来自哪里,因为Object是普通的java bean(但是使用继承)。

我的抽象类是:

@JsonSerialize(include=JsonSerialize.Inclusion.NON_NULL)
@JsonIgnoreProperties(ignoreUnknown = true)
@JsonTypeInfo(use=JsonTypeInfo.Id.CLASS, include=JsonTypeInfo.As.PROPERTY, property="@javaClass")
public abstract class AbstractBean implements Serializable {
    @JsonIgnore public static final transient IMarkupParser MARKUP_PARSER = new WikiMarkupParser();

    @JsonProperty("id")
    private String id;

    @JsonProperty("@class")
    private String clazz;

    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    public String getClazz() {
        return this.getClass().getSimpleName();
    }
}

我的孩子是:

public class Posting extends AbstractBean {
    private String postingSource;
    private String languageCode;

    public String getPostingSource() {
        return postingSource;
    }

    public void setPostingSource(String postingSource) {
        this.postingSource = postingSource;
    }

    public String getLanguageCode() {
        return languageCode;
    }

    public void setLanguageCode(String languageCode) {
        this.languageCode = languageCode;
    }
}

我不知道为什么serailizer甚至会尝试序列化mixin,因为它们不是bean的一部分但是在这里它们是(是的,我试图让它们也可序列化,就像测试一样,没有运气):< / p>

public interface IdMixins extends Serializable {
}

public interface MongoIdMixIn extends IdMixins {
    @JsonProperty("_id")
    @JsonSerialize(using = MongoIdSerializer.class)
    public String getId();

    @JsonProperty("_id")
    @JsonDeserialize(using = MongoIdDeserializer.class)
    public void setId(String id);
}

public class MongoIdDeserializer extends JsonDeserializer<String> implements Serializable {
    private static final long serialVersionUID = -5404276857799190647L;

    @Override
    public String deserialize(JsonParser jp, DeserializationContext deserializationContext) throws IOException, JsonProcessingException {
        String value = null;

        String tmp = jp.getText(); // {
        validate(jp, tmp,"{");
        int curly = 1;

        while (jp.nextToken() != null) {
            String v = jp.getText();

            if (v.equals("{")) curly++;

            if (v.equals("$oid")) {
                jp.nextToken();
                value = jp.getText();
            }

            if (v.equals("}")) curly--;
            if (curly<=0) return value;
        }

        return null;
    }

    private void validate(JsonParser jsonParser, String input, String expected) throws JsonProcessingException {
        if (!input.equals(expected)) {
            throw new JsonParseException("Unexpected token: " + input, jsonParser.getTokenLocation());
        }
    }
}

public class MongoIdSerializer extends JsonSerializer<String> implements Serializable {
    private static final long serialVersionUID = 3435689991839324194L;

    @Override
    public void serialize(String s, JsonGenerator jsonGenerator, SerializerProvider serializerProvider) throws IOException, JsonProcessingException {
        jsonGenerator.writeStartObject();
        jsonGenerator.writeFieldName("$oid");
        jsonGenerator.writeString(s);
        jsonGenerator.writeEndObject();
    }
}

2 个答案:

答案 0 :(得分:1)

愚蠢的我!序列化链中的某个地方是一个完全不必要的ObjectMapper对象。但很难找到,因为Posting对象不是真正的原因,而是另一个对象。但Stacktrace和com.fasterxml.jackson.databind.deser.impl.TypeWrappedDeserializer异常完全错过领先! ...集群软件有时候很难调试: - )

答案 1 :(得分:0)

我是1 Rep。指出能否发表评论。所以我必须提出一个建议作为答案;-)。

也许其中一个Annotations将TypeWrappedDeserializer的实例作为私有属性注入AbstractBean。也许是对反序列化机制的暗示。

您可以使用反射检查创建的对象以进行验证吗?

for (Field field : tgtObject.getClass().getDeclaredFields() )
{
   // you can replace this by your logging method
   System.out.println("Field: " + field.getName() + ":" + field.getType());
}
for (Field field : tgtObject.getClass().getSuperclass().getDeclaredFields() )
{
   // you can replace this by your logging method
   System.out.println("Field: " + field.getName() + ":" + field.getType());
}

如果您在列表中找到适当的类型,则通过字节代码增强添加类。