Jersey POJOMappingFeature从JSON中排除空属性

我已经使用Jersey Rest API和POJOMappingFeature将对象转换为JSON

<servlet>
    <servlet-name>Jersey Web Application</servlet-name>
    <servlet-class>com.sun.jersey.spi.spring.container.servlet.SpringServlet</servlet-class>
    <init-param>
        <param-name>com.sun.jersey.config.property.packages</param-name>
        <param-value>com.xxx.rest.api</param-value>
    </init-param>
    <init-param>
        <param-name>com.sun.jersey.api.json.POJOMappingFeature</param-name>
        <param-value>true</param-value>
    </init-param>
    <load-on-startup>1</load-on-startup>
</servlet>
<servlet-mapping>
    <servlet-name>Jersey Web Application</servlet-name>
    <url-pattern>/*</url-pattern>
</servlet-mapping>**

从我的Jersey REST API返回时，不会忽略null值。这意味着不应用JsonInclude.Include.NON_null

---

有什么想法吗？

@JsonInclude是来自的注释。POJOMappingFeature是一种Jersey功能，因此在序列化POJO时不会考虑Jackson注释

你可以用。然后，您的RESTAPI将能够使用Jackson将端点返回的对象序列化为JSON

@JsonInclude(JsonInclude.Include.NON_NULL)
public class User{
// Other properties goes here
}