java - 在配置全局设置的特殊情况下构建 Jacksons ObjectMapper

Question

在我的应用程序中，我将 Jackson 配置为全局使用 SerializationFeature.WRAP_ROOT_VALUE 和 DeserializationFeature.UNWRAP_ROOT_VALUE。

@Configuration
public class AppConfig {

    public Jackson2ObjectMapperBuilder jacksonBuilder() {
        Jackson2ObjectMapperBuilder builder = new Jackson2ObjectMapperBuilder();
        builder.featuresToEnable(SerializationFeature.WRAP_ROOT_VALUE, DeserializationFeature.UNWRAP_ROOT_VALUE);
        return builder;
    }
}

此配置工作正常，但现在我遇到的情况是，在反序列化情况下，我得到一个没有rootname的JSON响应。所以我有一个服务类，它使用 RestTemplateBuilder 构建 RestTemplate 并将一些数据POST 到 REST-Webservice。

@Service
public class ApiServiceImpl
        implements ApiService<RegisterResponse> {

    private RestTemplate restTemplate;

    public ApiServiceImpl(RestTemplateBuilder restTemplateBuilder) {
        restTemplate = restTemplateBuilder
                .errorHandler(new RestTemplateResponseErrorHandler()).build();
    }

    @Override
    public ResponseEntity<RegisterResponse> callAPI(String requestAsJson,
            String username, String password) {
        ResponseEntity<RegisterResponse> result = null;
        HttpHeaders headers = getHeaders(username, password);

        result = restTemplate.exchange(uri, HttpMethod.POST,
                new HttpEntity<String>(requestAsJson, headers),
                RegisterResponse.class);

        return result;
    }
}

响应如下所示:

{
    "redirect-url": "https://any-url.com/?with=params"
}

我想直接将其反序列化为以下 POJO。 (就像在restTemplate.exchange中配置的那样)

public class RegisterResponse {

    @JsonProperty("redirect-url")
    private String redirectUrl;
    //getter/setter
}

很明显，由于 UNWRAP_ROOT_VALUE 功能而出现此异常:

com.fasterxml.jackson.databind.exc.MismatchedInputException: Root name 'redirect-url' does not match expected ('RegisterResponse') for type [simple type, class xxx.xxx.xxxservice.xxx.model.response.entity.RegisterResponse]
 at [Source: (String)"{
    "redirect-url": "https://any-url.com/?with=params"
}"; line: 2, column: 5]
    at com.fasterxml.jackson.databind.exc.MismatchedInputException.from(MismatchedInputException.java:59)
    at com.fasterxml.jackson.databind.DeserializationContext.reportInputMismatch(DeserializationContext.java:1356)
    at com.fasterxml.jackson.databind.ObjectMapper._unwrapAndDeserialize(ObjectMapper.java:4087)

如何配置 Jackson 在这种特殊情况下不使用 DeserializationFeature.UNWRAP_ROOT_VALUE？

Answer 1

就像 JB Nizet 评论的那样，可能是通过将 Jakson 的 MappingJackson2HttpMessageConverter 和 ObjectMapper 的新实例设置到 MessageConverters 列表中来实现的。

restTemplate.getMessageConverters().add(getCustomConverter());

private MappingJackson2HttpMessageConverter getCustomConverter() {
        ObjectMapper mapper = new ObjectMapper();
        mapper.enable(DeserializationFeature.UNWRAP_ROOT_VALUE);
        mapper.disable(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES);

        MappingJackson2HttpMessageConverter customConverter = 
            new MappingJackson2HttpMessageConverter(mapper);
        if (!restTemplate.getMessageConverters()
                .removeIf(MappingJackson2HttpMessageConverter.class::isInstance)) {
            new RuntimeException("Custom MappingJackson2HttpMessageConverter not found");
        }
        return customConverter;
    }