在不使用dynamodbmarshalling的情况下在@DynamoDBTable中创建List <myobject>的替代方法(不建议使用)

时间:2018-02-08 15:16:06

标签: java spring jackson amazon-dynamodb

我通过创建 MyCustomMarshaller 跟随here

MyCustomMarshaller

public class MyCustomMarshaller implements DynamoDBMarshaller<List<DemoClass>> {

    private static final ObjectMapper mapper = new ObjectMapper();
    private static final ObjectWriter writer = mapper.writer();

    @Override
    public String marshall(List<DemoClass> obj) {

        try {
            return writer.writeValueAsString(obj);
        } catch (JsonProcessingException e) {
            throw failure(e,
                    "Unable to marshall the instance of " + obj.getClass()
                            + "into a string");
        }
    }

    @Override
    public List<DemoClass> unmarshall(Class<List<DemoClass>> clazz, String json) {
        final CollectionType
                type =
                mapper.getTypeFactory().constructCollectionType(List.class, DemoClass.class);
        try {
            return mapper.readValue(json, type);
        } catch (Exception e) {
            throw failure(e, "Unable to unmarshall the string " + json
                    + "into " + clazz);
        }
    }
}

我的dynamoDb类

@DynamoDBAttribute
@DynamoDBMarshalling(marshallerClass = MyCustomMarshaller.class)
List<DemoClass> Object;

DemoClass

public class DemoClass {

    String name;

    int id;

}

所有代码都运行良好。事情是

  

com.amazonaws.services.dynamodbv2.datamodeling.DynamoDBMarshalling是   的弃用

那么如何在不使用此dynamoDBmarshalling的情况下更改代码?

先谢谢,

2 个答案:

答案 0 :(得分:0)

是的,您应该使用DynamoDBTypeConverter

您可以从here

复制我的代码

为了完整性,这里是我在链接答案中使用的示例

// Model.java
@DynamoDBTable(tableName = "...")
public class Model {
  private String id;
  private List<MyObject> objects;

  public Model(String id, List<MyObject> objects) {
    this.id = id;
    this.objects = objects;
  }

  @DynamoDBHashKey(attributeName = "id")
  public String getId() { return this.id; }
  public void setId(String id) { this.id = id; }

  @DynamoDBTypeConverted(converter = MyObjectConverter.class)
  public List<MyObject> getObjects() { return this.objects; }
  public void setObjects(List<MyObject> objects) { this.objects = objects; }
}

-

public class MyObjectConverter implements DynamoDBTypeConverter<String, List<MyObject>> {

    @Override
    public String convert(List<Object> objects) {
        //Jackson object mapper
        ObjectMapper objectMapper = new ObjectMapper();
        try {
            String objectsString = objectMapper.writeValueAsString(objects);
            return objectsString;
        } catch (JsonProcessingException e) {
            //do something
        }
        return null;
    }

    @Override
    public List<Object> unconvert(String objectssString) {
        ObjectMapper objectMapper = new ObjectMapper();
        try {
            List<Object> objects = objectMapper.readValue(objectsString, new TypeReference<List<Object>>(){});
            return objects;
        } catch (JsonParseException e) {
            //do something
        } catch (JsonMappingException e) {
            //do something
        } catch (IOException e) {
            //do something
        }
        return null;
    }
}

答案 1 :(得分:0)

另一种更易于实现的方法,让DynamoDB在幕后处理转换。

将字段注释为@DynamoAttribute

@DynamoDBAttribute
private MyObjectClass myObject;

然后,您用@DynamoDBDocument注释“ MyObjectClass”

@DynamoDBDocument
public class MyObjectClass {
....
}

并且DynamoDB会将“ MyObjectClass myObject”转换和取消转换为JSON形状,即使它是对象的列表/数组也是如此。