用模块初始化Jersey客户端的正确方法是什么?
所以,这不应该这么难。 我正在尝试在Jersey客户端应用程序中使用ObjectMapper
反序列化java.time.LocalDateTime
。 可悲的是,这导致了例外:
Exception in thread "main" javax.ws.rs.client.ResponseProcessingException: com.fasterxml.jackson.databind.JsonMappingException: Can not construct instance of java.time.LocalDateTime: no suitable constructor found, can not deserialize from Object value (missing default constructor or creator, or perhaps need to add/enable type information?)
好的,所以有一个模块为Java时间types实现适当的序列化器/反序列化器: com.fasterxml.jackson.datatype.jsr310.JavaTimeModule
。 太棒了,所以我们只需要在应用程序中安装该模块,一切都很好。 但是,怎么样?
这个文件是沉默的。 我已经尝试了以下几行init代码的所有明智的组合,但没有运气:
ClientBuilder builder = ClientBuilder.newBuilder(); builder.register(new RequestFilter(this)); builder.register(new ResponseFilter(this)); builder.register(new JacksonJsonProvider(objectMapper)); builder.register(new JavaTimeModule()); ClientConfig cc = new ClientConfig(); cc.register(new JacksonJsonProvider(objectMapper)); cc.register(new JavaTimeModule()); // cc.getClasses().add(JavaTimeModule.class); // no go, this collection is unmodifiable builder.register(cc); // Client client = builder.build(); Client client = builder.withConfig(cc).build(); client.register(new JacksonJsonProvider(objectMapper)); client.register(new JavaTimeModule()); //client.getConfiguration().getClasses().add(JavaTimeModule.class); // and this one too.
上面的ObjectMapper
是一个单独的,在我的应用程序中放置任何人(用于客户端之外),并初始化为:
objectMapper = new ObjectMapper(); objectMapper.setSerializationInclusion(JsonInclude.Include.NON_NULL); objectMapper.configure(SerializationFeature.INDENT_OUTPUT, true); objectMapper.configure(SerializationFeature.WRITE_NULL_MAP_VALUES, false); objectMapper.registerModule(new JavaTimeModule());
它也不能映射LocalDateTime。
那么,配置Jersey客户端以使用该模块并反序列化java.time.LocalDateTime
的正确方法是什么?
尝试val mapper = ObjectMapper()mapper.registerModule(JavaTimeModule())
val provider = JacksonJaxbJsonProvider() provider.setMapper(mapper) val clientConfig = ClientConfig() clientConfig.register(provider)
(为kotlin道歉 – 按照您的预期转换为Java)。