json响应的样本如下所示:
{"publicId":"123","status":null,"partner":null,"description":null}
最好截断响应中的所有空对象。在这种情况下,响应将变为{"publicId":"123"}
。
有什么建议吗?谢谢!
P.S:我想我可以在泽西岛做到这一点。另外我相信他们都使用Jackson作为JSON处理器。后来添加: 我的配置:
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd">
<!-- Scans the classpath of this application for @Components to deploy as beans -->
<context:component-scan base-package="com.SomeCompany.web" />
<!-- Application Message Bundle -->
<bean id="messageSource" class="org.springframework.context.support.ReloadableResourceBundleMessageSource">
<property name="basename" value="/WEB-INF/messages/messages" />
<property name="cacheSeconds" value="0" />
</bean>
<!-- Configures Spring MVC -->
<import resource="mvc-config.xml" />
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:mvc="http://www.springframework.org/schema/mvc"
xsi:schemaLocation="
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/mvc http://www.springframework.org/schema/mvc/spring-mvc-3.0.xsd">
<!-- Configures the @Controller programming model -->
<mvc:annotation-driven />
<!-- Forwards requests to the "/" resource to the "welcome" view -->
<!--<mvc:view-controller path="/" view-name="welcome"/>-->
<!-- Configures Handler Interceptors -->
<mvc:interceptors>
<!-- Changes the locale when a 'locale' request parameter is sent; e.g. /?locale=de -->
<bean class="org.springframework.web.servlet.i18n.LocaleChangeInterceptor" />
</mvc:interceptors>
<!-- Saves a locale change using a cookie -->
<bean id="localeResolver" class="org.springframework.web.servlet.i18n.CookieLocaleResolver" />
<!--<bean class="org.springframework.web.servlet.view.ContentNegotiatingViewResolver">
<property name="mediaTypes">
<map>
<entry key="atom" value="application/atom+xml"/>
<entry key="html" value="text/html"/>
<entry key="json" value="application/json"/>
<entry key="xml" value="text/xml"/>
</map>
</property>
<property name="viewResolvers">
<list>
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/views/"/>
<property name="suffix" value=".jsp"/>
</bean>
</list>
</property>
<property name="defaultViews">
<list>
<bean class="org.springframework.web.servlet.view.json.MappingJacksonJsonView" />
<bean class="org.springframework.web.servlet.view.xml.MarshallingView" >
<property name="marshaller">
<bean class="org.springframework.oxm.jaxb.Jaxb2Marshaller" />
</property>
</bean>
</list>
</property>
</bean>
-->
<!-- Resolves view names to protected .jsp resources within the /WEB-INF/views directory -->
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/views/"/>
<property name="suffix" value=".jsp"/>
</bean>
我的代码:
@Controller
public class SomeController {
@RequestMapping(value = "/xyz", method = {RequestMethod.GET, RequestMethod.HEAD},
headers = {"x-requested-with=XMLHttpRequest","Accept=application/json"}, params = "!closed")
public @ResponseBody
List<AbcTO> getStuff(
.......
}
}
答案 0 :(得分:32)
是的,您可以通过使用@JsonSerialize(include=JsonSerialize.Inclusion.NON_NULL)
对其进行注释来为单个类执行此操作,或者您可以通过配置ObjectMapper
,将序列化包含设置为JsonSerialize.Inclusion.NON_NULL
来全面执行此操作。< / p>
以下是杰克逊常见问题解答中的一些信息:http://wiki.fasterxml.com/JacksonAnnotationSerializeNulls。
注释类很简单,但配置ObjectMapper
序列化配置稍微复杂一些。有关后者here的一些具体信息。
答案 1 :(得分:18)
不回答问题,但这是谷歌的第二个结果。
如果有人来到这里并且希望在春季4(就像它发生在我身上)那样做,你可以使用注释
@JsonInclude(Include.NON_NULL)
返回班级。
正如评论中所提到的,如果有人感到困惑,那么应该在将转换为JSON的类中使用注释。
答案 2 :(得分:9)
如果使用Spring Boot for REST,您可以在application.properties
:
spring.jackson.serialization-inclusion=NON_NULL
答案 3 :(得分:1)
上面的Java配置。只需将以下内容放在@Configuration类中即可。
@Bean
public Jackson2ObjectMapperBuilder objectMapperBuilder() {
Jackson2ObjectMapperBuilder builder = new Jackson2ObjectMapperBuilder();
builder.serializationInclusion(JsonInclude.Include.NON_NULL);
return builder;
}