Question

我在SpringMVC应用程序中使用MappingJacksonJsonView来从我的控制器渲染JSON。我希望我的对象中的ObjectId呈现为.toString，而是将ObjectId序列化为其部分。它在我的Velocity / JSP页面中运行得很好：

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中获取那么多信息？我只想要.toString（）方法，而不是所有细节。

感谢。

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;
    }
}

我的控制员：

@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";
    }
}

Answer 1

我不得不让getId（）方法返回一个String。这是让杰克逊停止序列化ObjectId的唯一方法。

public String getId() {
    if (id != null) {
        return id.toString();
    } else {
        return null;
    }
}

public void setId(ObjectId id) {
    this.id = id;
}

setId（）仍然必须是ObjectId，因此Mongo（及其驱动程序）可以正确设置ID。

Answer 2

之前的回答确实很有效，但它很丑陋且没有经过深思熟虑 - 一个明确的解决方法来实际修复问题。

真正的问题是ObjectId反序列化为其组成部分。 MappingJacksonJsonView ObjectId看到它是什么，一个对象，并开始研究它。在JSON中看到的反序列化字段是组成ObjectId的字段。要停止此类对象的序列化/反序列化，您必须配置扩展ObjectMapper的CustomObjectMapper。

以下是CustomeObjectMapper：

public class CustomObjectMapper extends ObjectMapper {

    public CustomObjectMapper() {
        CustomSerializerFactory sf = new CustomSerializerFactory();
        sf.addSpecificMapping(ObjectId.class, new ObjectIdSerializer());
        this.setSerializerFactory(sf);
    }
}

以下是ObjectIdSerializer使用的CustomObjectMapper：

public class ObjectIdSerializer extends SerializerBase<ObjectId> {

    protected ObjectIdSerializer(Class<ObjectId> t) {
        super(t);
    }

    public ObjectIdSerializer() {
        this(ObjectId.class);
    }

    @Override
    public void serialize(ObjectId value, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonGenerationException {
        jgen.writeString(value.toString());
    }
}

以下是需要在@Configuration - 注释类中更改的内容：

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

您基本上是在告诉Jackson如何序列化/反序列化此特定对象。像魅力一样。

Answer 3

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，在这种情况下，它是字符串。用于创建 ThingRepository 的 Thing 实体类将如下所示：

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

    String name;
}

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

java spring MappingJacksonJsonView在mongodb ObjectId上没有做toString

3 个答案: