Gab*_*adu 23 spring android jackson
我正在利用Spring提供的这个很酷的东西:Spring RESTWebService(spring的版本是3).如果我从浏览器访问URL,我可以看到JSON响应,但是从客户端端点(Android应用程序)iIreceive这个错误消息:
Caused by: org.springframework.web.client.ResourceAccessException:
I/O error: Can not deserialize instance of MyObject out of START_ARRAY token
at [Source: org.apache.http.conn.EofSensorInputStream@4076e940; line: 1,
column: 1]; nested exception is org.codehaus.jackson.map.JsonMappingException:
Can not deserialize instance of MyObject out of START_ARRAY token
at [Source: org.apache.http.conn.EofSensorInputStream@4076e940; line: 1, column: 1]
at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:466)
at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:414)
at org.springframework.web.client.RestTemplate.getForEntity(RestTemplate.java:234)
at com.be.android.locateconsultants.resources.AsyncTaskRESTServiceCaller.doInBackground(AsyncTaskRESTServiceCaller.java:43)
at com.be.android.locateconsultants.resources.AsyncTaskRESTServiceCaller.doInBackground(AsyncTaskRESTServiceCaller.java:1)
at android.os.AsyncTask$2.call(AsyncTask.java:252)
at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:305)
... 4 more
Caused by: org.codehaus.jackson.map.JsonMappingException: Can not deserialize
instance of MyObject out of START_ARRAY token
at [Source: org.apache.http.conn.EofSensorInputStream@4076e940; line: 1, column: 1]
at org.codehaus.jackson.map.deser.StdDeserializationContext.mappingException(StdDeserializationContext.java:198)
at org.codehaus.jackson.map.deser.BeanDeserializer.deserializeUsingCreator(BeanDeserializer.java:565)
at org.codehaus.jackson.map.deser.BeanDeserializer.deserialize(BeanDeserializer.java:365)
at org.codehaus.jackson.map.ObjectMapper._readMapAndClose(ObjectMapper.java:2395)
at org.codehaus.jackson.map.ObjectMapper.readValue(ObjectMapper.java:1655)
at org.springframework.http.converter.json.MappingJacksonHttpMessageConverter.readInternal(MappingJacksonHttpMessageConverter.java:135)
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$ResponseEntityResponseExtractor.extractData(RestTemplate.java:632)
at org.springframework.web.client.RestTemplate$ResponseEntityResponseExtractor.extractData(RestTemplate.java:618)
at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:459)
... 10 more
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MyObject结构与服务器端应用程序的结构相同.
我试图像这样请求服务器:
final String url = ".....";
RestTemplate restTemplate = new RestTemplate();
ResponseEntity<Consultant> responseEntity = restTemplate.getForEntity(
url, Consultant.class);
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或者像这样:
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_JSON);
HttpEntity<String> entity = new HttpEntity<String>(headers);
ResponseEntity<MyObject> response = restTemplate
.exchange("....",HttpMethod.GET, entity, MyObject.class);
System.out.println("RESPONSE: " + response.getBody());
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但仍然与上面相同的错误.无法弄清楚我在这一点上缺少什么,任何想法或提示都会很棒.谢谢.
paw*_*eba 23
你应该把它映射到List<Consultant>
而不是Consultant.class
.
这与Jackson以及您尝试从阵列反序列化和初始化容器的方式有关.
虽然我的使用上下文有点不同,但这可能有助于一些人从这里搜索特定于Jackson的反序列化错误.
我不得不这样做:
List<Consultant> consultants = Arrays.asList(
mapper.readValue(myJson.toString(), Consultants[].class)
);
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