我在我的 SpringMVC 应用程序中使用 MappingJacksonJsonView 从我的 Controller 呈现 JSON。我希望我的对象中的 ObjectId 呈现为 .toString 但它会将 ObjectId 序列化为其部分。它在我的 Velocity/JSP 页面中工作得很好:

Velocity:
    $thing.id
Produces:
    4f1d77bb3a13870ff0783c25


Json:
    <script type="text/javascript">
         $.ajax({
             type: 'GET',
             url: '/things/show/4f1d77bb3a13870ff0783c25',
             dataType: 'json',
             success : function(data) {
                alert(data);
             }
         });
    </script>
Produces:
    thing: {id:{time:1327331259000, new:false, machine:974358287, timeSecond:1327331259, inc:-260555739},…}
        id: {time:1327331259000, new:false, machine:974358287, timeSecond:1327331259, inc:-260555739}
            inc: -260555739
            machine: 974358287
            new: false
            time: 1327331259000
            timeSecond: 1327331259
        name: "Stack Overflow"


XML:
    <script type="text/javascript">
         $.ajax({
             type: 'GET',
             url: '/things/show/4f1d77bb3a13870ff0783c25',
             dataType: 'xml',
             success : function(data) {
                alert(data);
             }
         });
    </script>
Produces:
    <com.place.model.Thing>
        <id>
            <__time>1327331259</__time>
            <__machine>974358287</__machine>
            <__inc>-260555739</__inc>
            <__new>false</__new>
        </id>
        <name>Stack Overflow</name>
    </com.place.model.Thing>

有没有办法阻止 MappingJacksonJsonView 从 ObjectId 获取那么多信息？我只想要 .toString() 方法，而不是所有细节。

谢谢。

添加 Spring 配置:

@Configuration
@EnableWebMvc
public class MyConfiguration {

    @Bean(name = "viewResolver")
    public ContentNegotiatingViewResolver viewResolver() {
        ContentNegotiatingViewResolver contentNegotiatingViewResolver = new ContentNegotiatingViewResolver();
        contentNegotiatingViewResolver.setOrder(1);
        contentNegotiatingViewResolver.setFavorPathExtension(true);
        contentNegotiatingViewResolver.setFavorParameter(true);
        contentNegotiatingViewResolver.setIgnoreAcceptHeader(false);
        Map<String, String> mediaTypes = new HashMap<String, String>();
        mediaTypes.put("json", "application/x-json");
        mediaTypes.put("json", "text/json");
        mediaTypes.put("json", "text/x-json");
        mediaTypes.put("json", "application/json");
        mediaTypes.put("xml", "text/xml");
        mediaTypes.put("xml", "application/xml");
        contentNegotiatingViewResolver.setMediaTypes(mediaTypes);
        List<View> defaultViews = new ArrayList<View>();
        defaultViews.add(xmlView());
        defaultViews.add(jsonView());
        contentNegotiatingViewResolver.setDefaultViews(defaultViews);
        return contentNegotiatingViewResolver;
    }

    @Bean(name = "xStreamMarshaller")
    public XStreamMarshaller xStreamMarshaller() {
        return new XStreamMarshaller();
    }

    @Bean(name = "xmlView")
    public MarshallingView xmlView() {
        MarshallingView marshallingView = new MarshallingView(xStreamMarshaller());
        marshallingView.setContentType("application/xml");
        return marshallingView;
    }

    @Bean(name = "jsonView")
    public MappingJacksonJsonView jsonView() {
        MappingJacksonJsonView mappingJacksonJsonView = new MappingJacksonJsonView();
        mappingJacksonJsonView.setContentType("application/json");
        return mappingJacksonJsonView;
    }
}

还有我的 Controller :

@Controller
@RequestMapping(value = { "/things" })
public class ThingController {

    @Autowired
    private ThingRepository thingRepository;

    @RequestMapping(value = { "/show/{thingId}" }, method = RequestMethod.GET)
    public String show(@PathVariable ObjectId thingId, Model model) {
        model.addAttribute("thing", thingRepository.findOne(thingId));
        return "things/show";
    }
}

最佳答案

默认情况下，Jackson 提供接收到的对象的序列化。 ObjectId 返回对象，因此其属性在转换为 JSON 后可见。您需要指定所需的序列化类型，在本例中为字符串。用于创建 ThingRepository 的 Thing 实体类将如下所示来完成此操作:

public class Thing {
    @Id
    @JsonSerialize(using= ToStringSerializer.class)
    ObjectId id;

    String name;
}

这里记下添加的注释 @JsonSerialize(using= ToStringSerializer.class) 指示将 ObjectID 序列化为 String。

关于java spring MappingJacksonJsonView 没有在 mongodb ObjectId 上执行 toString，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/9065042/