java - JSON : Unrecognized field "value" (), 未标记为可忽略

Question

谁能帮我弄清楚需要添加什么？

JSON:

{"value":{"keyword":"better","correct":"","page":0,"size":10,"cost":51,"total":1107}}

项目等级

@JsonAutoDetect
@JsonSerialize(include = Inclusion.NON_NULL)
@JsonRootName(value = "value")    
public class Response {

private int page;
private int size;
private int total;
private int cost;
private int result;

private String keyword;
private String correct;

它仍然得到“Servlet.service() for servlet appServlet throw exception”

org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "value" (), not marked as ignorable"

Answer 1

尝试将此添加到您的映射器配置

mapper.configure(SerializationConfig.Feature.WRAP_ROOT_VALUE, true);
mapper.configure(DeserializationConfig.Feature.UNWRAP_ROOT_VALUE, true);

如果您使用 RestTemplate，您将需要配置底层 jackson 映射器。您可以通过配置映射器并将其设置在转换器中来完成此操作。请参阅下面的代码。

ObjectMapper mapper = new ObjectMapper();
mapper.configure(SerializationConfig.Feature.WRAP_ROOT_VALUE, true);
mapper.configure(DeserializationConfig.Feature.UNWRAP_ROOT_VALUE, true);


MappingJacksonHttpMessageConverter messageConverter = new MappingJacksonHttpMessageConverter();
messageConverter.setObjectMapper(mapper);
List<HttpMessageConverter<?>> messageConverters = new ArrayList<HttpMessageConverter<?>>();
messageConverters.add(messageConverter);
RestTemplate restTemplate = new RestTemplate();
restTemplate.setMessageConverters(messageConverters);

查看此处了解更多详情:https://jira.springsource.org/browse/ANDROID-45