json - 如何使用 Jackson 使用多参数构造函数序列化 scala 对象？

Question

class AgentResponse[T] @JsonCreator()(@JsonProperty("result") val result: T, @JsonProperty("status") val status: ResponseStatus)

class ResponseStatus @JsonCreator()(@JsonProperty("succeeded") val succeeded: Boolean, @JsonProperty("message") val message: String, @JsonProperty("timeStamp") val timeStamp: Long)

new ObjectMapper().registerModule(DefaultScalaModule).writer().writeValue(out, new AgentResponse(result, new ResponseStatus(true, "OK", now)))

它抛出错误:

JsonMappingException: No serializer found for class com.fg.mail.smtp.rest.Handler$AgentResponse and no properties discovered to create BeanSerializer (to avoid exception, disable SerializationConfig.SerializationFeature.FAIL_ON_EMPTY_BEANS) )

scala 对象应该如何使其按预期工作？

Answer 1

引自 scala.annotation.meta :

By default, annotations on (val-, var- or plain) constructor parameters end up on the parameter, not on any other entity. Annotations on fields by default only end up on the field.

The meta-annotations in package scala.annotation.meta are used to control where annotations on fields and class parameters are copied. This is done by annotating either the annotation type or the annotation class with one or several of the meta-annotations in this package.

所以注释转到构造函数参数。您需要将它分配给 getter 以及构造函数参数:

class MyClass @JsonCreator() (@(JsonProperty @getter @param) val arg: String)