本文介绍了jackson为什么我需要在子类上使用JsonTypeName注释的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
此链接
我正在尝试理解为什么我(可能)需要 @JsonTypeName 关于子类(例如所有'互联网; sujests to put )如果它没有它?
I'm trying to understand why do I (may) need @JsonTypeName on subclasses (like all 'internet; sujests to put) if it works without it ?
@JsonTypeInfo(use = JsonTypeInfo.Id.NAME, property = "aType") @JsonSubTypes(Array( new Type(value = classOf[ModelA], name = "ModelA"), new Type(value = classOf[ModelB], name = "ModelB") )) class BaseModel(val modelName:String) //@JsonTypeName("SomeModel") // Commented. Do I need this? class ModelA(val a:String, val b:String, val c:String, commonData:String) extends BaseModel(commonData) { def this() = this("default", "default", "default" ,"default") } //@JsonTypeName("SomeModel") // Commented. Do I need this? class ModelB(val a:String, val b:String, val c:String, commonData:String) extends BaseModel(commonData) { def this() = this("default", "default", "default" ,"default") }推荐答案
你不需要它们。
文档 @JsonSubTypes.Type 解释
子类型的定义以及可选名称。如果缺少name,将检查类型的类是否为JsonTypeName注释;如果它也缺失或为空,则默认名称将由类型id机制构造。默认名称通常基于班级名称。
Definition of a subtype, along with optional name. If name is missing, class of the type will be checked for JsonTypeName annotation; and if that is also missing or empty, a default name will be constructed by type id mechanism. Default name is usually based on class name.
你应该有
@JsonSubTypes(Array( new Type(value = classOf[ModelA], name = "ModelA") ... class ModelA或
@JsonSubTypes(Array( new Type(value = classOf[ModelA]) ... @JsonTypeName("ModelA") class ModelA更多推荐
jackson为什么我需要在子类上使用JsonTypeName注释
发布评论