使用RestTemplate反序列化嵌套对象

Question

我正在使用RestTemplate并解决对象的反序列化问题.这就是我在做的事情.JSON响应看起来像,

{
"response": {
"Time": "Wed 2013.01.23 at 03:35:25 PM UTC",
"Total_Input_Records": 5,
},-
"message": "Succeeded",
"code": "200"
}

使用jsonschema2pojo将此Json有效负载转换为POJO

public class MyClass {
    @JsonProperty("response")
    private Response response;
    @JsonProperty("message")
    private Object message;
    @JsonProperty("code")
    private Object code;
    private Map<String, Object> additionalProperties = new HashMap<String, Object>();
    //bunch of getters and setters here
}
public class Response {
    @JsonProperty("Time")
    private Date Time;
    @JsonProperty("Total_Input_Records")
    private Object Total_Input_Records;
    private Map<String, Object> additionalProperties = new HashMap<String, Object>();
//bunch of getters and setters here
}

这是我获得异常的请求处理,

String url = "http://my.site.com/someparams";
RestTemplate template = new RestTemplate(
                new HttpComponentsClientHttpRequestFactory());
FormHttpMessageConverter converter = new FormHttpMessageConverter();
List<MediaType> mediaTypes = new ArrayList<MediaType>();
mediaTypes.add(new MediaType("application", "x-www-form-urlencoded"));
converter.setSupportedMediaTypes(mediaTypes);
template.getMessageConverters().add(converter);
MyClass upload = template.postForObject(url, null, MyClass.class);

这是令人沮丧的部分,例外(故意修剪,不完整).我错过了什么？

org.springframework.http.converter.HttpMessageNotReadableException: Could not read JSON: Unrecognized field "Time" (Class com.temp.pointtests.Response), not marked as ignorable
 at [Source: org.apache.http.conn.EofSensorInputStream@340ae1cf; line: 1, column: 22  (through reference chain: com.temp.pointtests.MyClass["response"]->com.temp.pointtests.Response["Time"]);] 

+++++更新解决了++++++

我看到Spring添加了使用Jackson 2的MappingJackson2HttpMessageConverter.因为我上面的代码中的MappingJacksonHttpMessageConverter使用的是Jackson Pre2.0版本,但它不起作用.但它适用于Jackson 2.0.现在可以使用MappingJackson2HttpMessageConverter,我现在可以将它添加到我的RestTemplate中,一切正常:-).以下是具有相同问题的人的代码,

String url = "http://mysite.com/someparams";
RestTemplate template = new RestTemplate();
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_FORM_URLENCODED);  
HttpEntity request = new HttpEntity(headers);
List<HttpMessageConverter<?>> messageConverters = new ArrayList<HttpMessageConverter<?>>();
MappingJackson2HttpMessageConverter map = new MappingJackson2HttpMessageConverter();
messageConverters.add(map);
messageConverters.add(new FormHttpMessageConverter());
template.setMessageConverters(messageConverters);
MyClass msg = template.postForObject(url, request, MyClass.class);

Answer 1

使用 org.codehaus.jackson.map.JsonDeserializer 的 @JsonSerialize(using = JsonDateSerializer.class) 或 @JsonDeserialize(using = JsonDateDeSerializer.class) 注解；它将解决问题或用户 ObjectMapper(org.codehaus.jackson.map.ObjectMapper) 转换为 Json 字符串。

objectMapper.writeValueAsString(Object); // 这将给出 json 字符串