Question

我正在使用一些遗留代码，我正在尝试编写一个lambda来处理函数。

该功能看起来像

public Task doTask(Message message) throws Exception {
   LOG.debug("debug message");
   // ... more code
}

但是，参数Message在不同的模块中定义（使用getter和setter）（并作为依赖项传入）。结果我得到了错误：

{
  "errorMessage": "An error occurred during JSON parsing",
  "errorType": "java.lang.RuntimeException",
  "stackTrace": [],
  "cause": {
    "errorMessage": "com.fasterxml.jackson.databind.JsonMappingException: Can not construct instance of com.mywebsite.messaging.Message, problem: abstract types either need to be mapped to concrete types, have custom deserializer, or be instantiated with additional type information\n at [Source: lambdainternal.util.NativeMemoryAsInputStream@31610302; line: 1, column: 1]",
    "errorType": "java.io.UncheckedIOException",
    "stackTrace": [],
    "cause": {
      "errorMessage": "Can not construct instance of com.mywebsite.messaging.Message, problem: abstract types either need to be mapped to concrete types, have custom deserializer, or be instantiated with additional type information\n at [Source: lambdainternal.util.NativeMemoryAsInputStream@31610302; line: 1, column: 1]",
      "errorType": "com.fasterxml.jackson.databind.JsonMappingException",
      "stackTrace": [
        "com.fasterxml.jackson.databind.JsonMappingException.from(JsonMappingException.java:148)",
        "com.fasterxml.jackson.databind.DeserializationContext.instantiationException(DeserializationContext.java:892)",
        "com.fasterxml.jackson.databind.deser.AbstractDeserializer.deserialize(AbstractDeserializer.java:139)",
        "com.fasterxml.jackson.databind.ObjectReader._bindAndClose(ObjectReader.java:1511)",
        "com.fasterxml.jackson.databind.ObjectReader.readValue(ObjectReader.java:1102)"
      ]
    }
  }
}

如何序列化这个甚至不在我模块中的对象？

非常感谢任何和所有帮助。

谢谢！

Answer 1

看起来com.mywebsite.messaging.Message是一个抽象类/ 接口。在这种情况下：使用@JsonDeserialize将解决问题。

这样的事情：

@JsonDeserialize(using = MessageDeserializer.class)
interface Message {
}

@JsonDeserialize(as = MessageImpl.class)
public class MessageImpl implements Message{
//write your implementation
}   

public class MessageDeserializer extends JsonDeserializer<Message> {

    @Override
    public Message deserialize(JsonParser jp, DeserializationContext context) throws IOException {
        ObjectMapper mapper = (ObjectMapper) jp.getCodec();
        ObjectNode root = mapper.readTree(jp);
        return mapper.readValue(root.toString(), MessageImpl.class);
    }
}

这些链接可以帮助您： How to add custom deserializer to interface using jackson http://www.baeldung.com/jackson-exception

希望这有帮助！

AWS Lambda在另一个模块中序列化对象

1 个答案: