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在调用服务时，我无法让 Jackson 正确地将 json 反序列化为对象（特别是我们正在使用 Jackson 使用 JAXB 注释的能力，因为我们还希望服务使用 XML）。我正在使用 Spring MVC，并且正在使用 RestTemplate 类来调用服务。

这是我为我的 junit 设置 MappingJacksonHttpMessageConverter 的地方：

ObjectMapper jsonMapper = new ObjectMapper();
AnnotationIntrospector introspector = new JaxbAnnotationIntrospector();
jsonMapper.getDeserializationConfig().setAnnotationIntrospector(introspector);
jsonMapper.getSerializationConfig().setAnnotationIntrospector(introspector);
jsonMapper.getSerializationConfig().setSerializationInclusion(Inclusion.NON_NULL);
MappingJacksonHttpMessageConverter jacksonConverter = new MappingJacksonHttpMessageConverter();
jacksonConverter.setObjectMapper(jsonMapper);
List<HttpMessageConverter<?>> converters = new ArrayList<HttpMessageConverter<?>>();
converters.add(jacksonConverter);
template.setMessageConverters(converters);

我这样称呼服务：

HttpHeaders requestHeaders = new HttpHeaders();
requestHeaders.set("Accept", "application/json");
HttpEntity<String> requestEntity = new HttpEntity<String>(requestHeaders);
ResponseEntity<NamedSystem> responseEntity = template.exchange(baseURL + "/{NamedSystemId}", 
        HttpMethod.GET, requestEntity, NamedSystem.class, orgId1);

我的NamedSystem班级是这样设置的：

@XmlRootElement(name = "NamedSystem", namespace = "http://schemas.abc.workplace.com/NamedSystem")
public class NamedSystem {
    private String id;
    private String name;
    private String description;
    private Set<NamedSystemAlias> aliases;
    private String href;

    @XmlAttribute(required = false, name = "id")
    public String getId() {
        return id;
    }


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


    @XmlAttribute(required = false, name = "name")
    public String getName() {
        return name;
    }


    public void setName(String name) {
        this.name = name;
    }


    @XmlAttribute(required = false, name = "description")
    public String getDescription() {
        return description;
    }


    public void setDescription(String description) {
        this.description = description;
    }


    @XmlElementWrapper(required = false, name = "aliases", namespace = "http://schemas.abc.workplace.com/NamedSystem")
    @XmlElement(required = false, name = "alias", namespace = "http://schemas.abc.workplace.com/NamedSystem")
    public Set<NamedSystemAlias> getAliases() {
        return aliases;
    }


    public void setAliases(Set<NamedSystemAlias> aliases) {
        this.aliases = aliases;
    }

    @XmlAttribute(required = true, name = "href")
    public String getHref() {
        return href;
    }


    public void setHref(String href) {
        this.href = href;
    }
}

这是导致的错误：

org.springframework.web.client.ResourceAccessException: I/O error: Unrecognized field "NamedSystem" (Class com.workplace.abc.named.NamedSystem), not marked as ignorable
 at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@c1429c; line: 1, column: 2]; nested exception is org.codehaus.jackson.map.JsonMappingException: Unrecognized field "NamedSystem" (Class com.workplace.abc.named.NamedSystem), not marked as ignorable
 at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@c1429c; line: 1, column: 2]
 at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:453)
....
Caused by: org.codehaus.jackson.map.JsonMappingException: Unrecognized field "NamedSystem" (Class com.workplace.abc.named.NamedSystem), not marked as ignorable
 at [Source: sun.net.www.protocol.http.HttpURLConnection$HttpInputStream@c1429c; line: 1, column: 2]
 at org.codehaus.jackson.map.JsonMappingException.from(JsonMappingException.java:159)
 at org.codehaus.jackson.map.deser.StdDeserializationContext.unknownFieldException(StdDeserializationContext.java:247)
 at org.codehaus.jackson.map.deser.StdDeserializer.reportUnknownProperty(StdDeserializer.java:366)
 at org.codehaus.jackson.map.deser.StdDeserializer.handleUnknownProperty(StdDeserializer.java:352)
 at org.codehaus.jackson.map.deser.BeanDeserializer.handleUnknownProperty(BeanDeserializer.java:543)
 at org.codehaus.jackson.map.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:402)
 at org.codehaus.jackson.map.deser.BeanDeserializer.deserialize(BeanDeserializer.java:287)
 at org.codehaus.jackson.map.ObjectMapper._readMapAndClose(ObjectMapper.java:1588)
 at org.codehaus.jackson.map.ObjectMapper.readValue(ObjectMapper.java:1172)
 at org.springframework.http.converter.json.MappingJacksonHttpMessageConverter.readInternal(MappingJacksonHttpMessageConverter.java:132)
 at org.springframework.http.converter.AbstractHttpMessageConverter.read(AbstractHttpMessageConverter.java:154)
 at org.springframework.web.client.HttpMessageConverterExtractor.extractData(HttpMessageConverterExtractor.java:74)
 at org.springframework.web.client.RestTemplate$ResponseEntityResponseExtractor.extractData(RestTemplate.java:619)
 at org.springframework.web.client.RestTemplate$ResponseEntityResponseExtractor.extractData(RestTemplate.java:1)
 at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:446)
 ... 32 more

它似乎无法识别能够反序列化的 rootElement 'NamedSystem'。我将如何让它做到这一点？我已经看到使用相同 JAXB 注释的示例并且它们工作正常，所以我不确定我的案例有什么不同，或者我可能如何强制它正确反序列化它。如果有人可以提供任何帮助，我将不胜感激。

4

1 回答 1

2

如果有人遇到这种问题，这可能会为您解决：Enable Jackson to not output the class name when serializing (using Spring MVC)

请参阅我的答案并按照链接获取示例。

于 2010-07-23T14:04:14.767 回答