Spring Boot 1.2.5 & Jersey 2 忽略空字段

Question

在带有 Jersey 2 接口的 Spring boot 1.2.5 中，如何将 JSON 编组器设置为不包含具有空值的字段？

例如：

[
    {
        "created": 1433987509174,
        "lastModified": 1433876475580,
        "id": 1,
        "example": "example1b"
    },
    {
        "created": 1434502031212,
        "lastModified": 1434502031212,
        "id": 10000,
        "example": "example1c"
    },
    {
        "created": 1439151444176,
        "lastModified": 1439151444176,
        "id": 10011,
        "example": null
    }
]

字段"example": null根本不应该包含在json输出中，但这里指定它为空。

在我的@SpringBootApplication 类中，我尝试添加：

@Bean
public MappingJackson2HttpMessageConverter mappingJackson2HttpMessageConverter() {
    final MappingJackson2HttpMessageConverter converter = new MappingJackson2HttpMessageConverter();
    final ObjectMapper objectMapper = new ObjectMapper();
    objectMapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);
    converter.setObjectMapper(objectMapper);
    return converter;
}

或

@Bean
@Primary
public Jackson2ObjectMapperBuilder objectMapperBuilder() {
    Jackson2ObjectMapperBuilder builder = new Jackson2ObjectMapperBuilder();
    builder.serializationInclusion(JsonInclude.Include.NON_NULL);
    return builder;
}

或

@Primary
@Bean
public ObjectMapper mapper() {
    final ObjectMapper objectMapper = new ObjectMapper();
    objectMapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);
    return objectMapper;
}

和/或将@JsonSerialize(include = Inclusion.NON_NULL) 添加到对象本身

但它仍然会产生与上述相同的响应，包括 "example": null。

它使用 @JsonSerialize(include=Inclusion.NON_NULL) 在 Spring 3.0.7 上工作，但现在我已经移植到 Spring Boot 1.2.5，它不再工作。

我相信我已经按照文档 http://docs.spring.io/spring-boot/docs/current/reference/html/howto-spring-mvc.html#howto-customize-the-jackson-objectmapper 进行操作，但它不起作用，所以我希望有人可能会看到我遗漏的东西？提前致谢！

编辑：也只是尝试添加类：

@Configuration
public class WebConfiguration extends WebMvcAutoConfiguration {
    @Primary
    @Bean
    public ObjectMapper mapper() {
        final ObjectMapper objectMapper = new ObjectMapper();
        objectMapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);
        return objectMapper;
    }
}

解决方案：

package com.my.spring;

import javax.ws.rs.ext.ContextResolver;

import org.glassfish.jersey.server.ResourceConfig;
import org.glassfish.jersey.servlet.ServletProperties;
import org.springframework.context.annotation.Configuration;

import com.my.spring.service.rs.MyRestServiceImpl;
import com.fasterxml.jackson.annotation.JsonInclude;
import com.fasterxml.jackson.databind.ObjectMapper;

@Configuration
public class JerseyConfiguration extends ResourceConfig {

    public class ObjectMapperContextResolver implements ContextResolver<ObjectMapper> {
        private final ObjectMapper mapper;

        public ObjectMapperContextResolver() {
            mapper = new ObjectMapper();
            mapper.setSerializationInclusion(JsonInclude.Include.NON_NULL);
        }

        @Override
        public ObjectMapper getContext(Class<?> type) {
           return mapper;
        }
    }

    public JerseyConfiguration() {
        register(new ObjectMapperContextResolver());
        register(MyRestServiceImpl.class); // My jax-rs implementation class
        property(ServletProperties.FILTER_FORWARD_ON_404, true); // Not needed for this non_null issue
    }
}

Answer 1

我不知道混合 Spring 方式（配置映射器）以及 Jersey 如何处理这个问题。但是泽西岛配置ObjectMapper 的方式是通过ContextResolver，如this answer 所示。

然后用your Jersey configuration注册ObjectMapperContextResolver。

 public JerseyConfig extends ResourceConfig {
     public JerseyConfig() {
         ...
         register(ObjectMapperContextResolver.class);
     }
 }

或者，如果您正在扫描包，@Provider 注释将选择该类。