我正在努力配置要由Spring Integration DSL转换器使用的“自定义”
ObjectMapper。我收到了java.time.Instant我想解析为对象属性的json表示形式。即：
{“ type”：“ TEST”，“ source”：“ TEST”，“ timestamp”：{“ epochSecond”：1454503381，“
nano”：335000000}}

该消息是一个kafka消息，引发了一个问题：我是否应该编写一个实现Kafka编码器/解码器的自定义序列化程序，以便能够将kafka消息转换为正确的对象，或者spring-
integration必须自动执行此操作？

固件/依赖关系和版本：Spring Boot-1.3.2.RELEASE
Spring Integration Java Dsl-1.1.1.RELEASE
FasterXml Jackson-2.6.5

我已根据Jackson文档将此Java配置添加到项目中：https : //github.com/FasterXML/jackson-datatype-
jsr310

@Configuration
public class IntegrationConfiguration {

    @Bean
    public JsonObjectMapper<JsonNode, JsonParser> jsonObjectMapper() {
        ObjectMapper mapper = new ObjectMapper();
        mapper.registerModule(new JavaTimeModule());
        return new Jackson2JsonObjectMapper(mapper);
    }
}

以及以下Jackson JSR-310人工制品：

<dependency>
    <groupId>com.fasterxml.jackson.datatype</groupId>
    <artifactId>jackson-datatype-jsr310</artifactId>
    <version>2.6.5</version>
</dependency>

根据Spring博客上的这篇文章，我什至不必注册新的Java8时间模块。
https://spring.io/blog/2014/12/02/latest-jackson-integration-improvements-in-
spring#jackson-modules

这是我得到的例外：

Caused by: com.fasterxml.jackson.databind.JsonMappingException: No suitable constructor found for type [simple type, class java.time.Instant]: can not instantiate from JSON object (missing default constructor or creator, or perhaps need to add/enable type information?)
 at [Source: {"type":"TEST","source":"TEST","timestamp":{"epochSecond":1454503381,"nano":335000000}}; line: 1, column: 71] (through reference chain: my.app.MyDto["timestamp"])
    at com.fasterxml.jackson.databind.JsonMappingException.from(JsonMappingException.java:148)
    at com.fasterxml.jackson.databind.deser.BeanDeserializerBase.deserializeFromObjectUsingNonDefault(BeanDeserializerBase.java:1106)
    at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:296)
    at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:133)
    at com.fasterxml.jackson.databind.deser.SettableBeanProperty.deserialize(SettableBeanProperty.java:520)
    at com.fasterxml.jackson.databind.deser.impl.MethodProperty.deserializeAndSet(MethodProperty.java:95)
    at com.fasterxml.jackson.databind.deser.BeanDeserializer.vanillaDeserialize(BeanDeserializer.java:258)
    at com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:125)
    at com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:3736)
    at com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:2764)
    at org.springframework.integration.support.json.Jackson2JsonObjectMapper.fromJson(Jackson2JsonObjectMapper.java:75)
    at org.springframework.integration.support.json.Jackson2JsonObjectMapper.fromJson(Jackson2JsonObjectMapper.java:44)
    at org.springframework.integration.support.json.AbstractJacksonJsonObjectMapper.fromJson(AbstractJacksonJsonObjectMapper.java:56)
    at org.springframework.integration.json.JsonToObjectTransformer.doTransform(JsonToObjectTransformer.java:78)
    at org.springframework.integration.transformer.AbstractTransformer.transform(AbstractTransformer.java:33)
    ... 74 more

解决方案：
问题是我希望Spring能够检测到jackson-datatype-
jsr310原型并注册JavaTimeModule，但事实并非如此。有两种方法可以解决此问题：
1.如果我们将Spring Boot与Spring Integration一起使用，则可以接受。
2.如果使用Spring Integration
Dsl，只需将IntegrationConfiguration类与jsonObjectMapper（）bean一起使用，并像这样使用它：

@Autowired
private JsonObjectMapper jsonObjectMapper;

return IntegrationFlows
        .from(inboundChannel())
        .transform(Transformers.fromJson(myDto.class, jsonObjectMapper))
        ...