我正在开发与我的服务器通信的Android应用程序。这种沟通是通过Spring框架和Jackson完成的。我正在成功发送我的服务器请求，但我没有收到回复。这就是我所做的：
Android应用：
public Loja getLoja() {
RestTemplate restTemplate = new RestTemplate();
restTemplate.getMessageConverters().add(new MappingJacksonHttpMessageConverter());
restTemplate.setRequestFactory(new HttpComponentsClientHttpRequestFactory());
String url = BASE_URL + "/android/played.json";
return restTemplate.getForObject(url, Loja.class);
}
Loja课程（我有两个版本。一个在Android应用程序中，另一个在Server中。它们完全相同）：
public class Loja {
private String nome;
private String xValue;
private String yValue;
private String andar;
public Loja(String nome, String xValue, String yValue, String andar) {
this.nome = nome;
this.xValue = xValue;
this.yValue = yValue;
this.andar = andar;
}
public Loja() {
}
public String getAndar() {
return andar;
}
public void setAndar(String andar) {
this.andar = andar;
}
public String getNome() {
return nome;
}
public void setNome(String nome) {
this.nome = nome;
}
public String getxValue() {
return xValue;
}
public void setxValue(String xValue) {
this.xValue = xValue;
}
public String getyValue() {
return yValue;
}
public void setyValue(String yValue) {
this.yValue = yValue;
}
}
这是我的控制器：
@RequestMapping("/android/played")
public ModelAndView getLoja() {
System.out.println("Android Resquest.");
Loja loja = new Loja("teste", "20", "30", "1");
ModelAndView mav = new ModelAndView();
mav.addObject("Loja", loja);
return mav;
}
通过所有这些，我在Android应用程序中获得了以下异常：
03-21 22:13:06.197: E/AndroidRuntime(25342): java.lang.RuntimeException: An error occured while executing doInBackground()
03-21 22:13:06.197: E/AndroidRuntime(25342): at android.os.AsyncTask$3.done(AsyncTask.java:299)
03-21 22:13:06.197: E/AndroidRuntime(25342): at java.util.concurrent.FutureTask$Sync.innerSetException(FutureTask.java:273)
03-21 22:13:06.197: E/AndroidRuntime(25342): at java.util.concurrent.FutureTask.setException(FutureTask.java:124)
03-21 22:13:06.197: E/AndroidRuntime(25342): at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:307)
03-21 22:13:06.197: E/AndroidRuntime(25342): at java.util.concurrent.FutureTask.run(FutureTask.java:137)
03-21 22:13:06.197: E/AndroidRuntime(25342): at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:230)
03-21 22:13:06.197: E/AndroidRuntime(25342): at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1076)
03-21 22:13:06.197: E/AndroidRuntime(25342): at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:569)
03-21 22:13:06.197: E/AndroidRuntime(25342): at java.lang.Thread.run(Thread.java:856)
03-21 22:13:06.197: E/AndroidRuntime(25342): Caused by: org.springframework.http.converter.HttpMessageNotReadableException: Could not read JSON: Unrecognized field "Loja" (Class com.example.androidspring.Loja), not marked as ignorable
03-21 22:13:06.197: E/AndroidRuntime(25342): at [Source: org.apache.http.conn.EofSensorInputStream@422e3c58; line: 1, column: 10] (through reference chain: com.example.androidspring.Loja["Loja"]); nested exception is org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "Loja" (Class com.example.androidspring.Loja), not marked as ignorable
03-21 22:13:06.197: E/AndroidRuntime(25342): at [Source: org.apache.http.conn.EofSensorInputStream@422e3c58; line: 1, column: 10] (through reference chain: com.example.androidspring.Loja["Loja"])
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.springframework.http.converter.json.MappingJacksonHttpMessageConverter.readInternal(MappingJacksonHttpMessageConverter.java:125)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.springframework.http.converter.AbstractHttpMessageConverter.read(AbstractHttpMessageConverter.java:147)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.springframework.web.client.HttpMessageConverterExtractor.extractData(HttpMessageConverterExtractor.java:76)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.springframework.web.client.RestTemplate.doExecute(RestTemplate.java:484)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.springframework.web.client.RestTemplate.execute(RestTemplate.java:439)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.springframework.web.client.RestTemplate.getForObject(RestTemplate.java:237)
03-21 22:13:06.197: E/AndroidRuntime(25342): at com.example.androidspring.MainActivity.getLoja(MainActivity.java:59)
03-21 22:13:06.197: E/AndroidRuntime(25342): at com.example.androidspring.MainActivity$DownloadFilesTask.doInBackground(MainActivity.java:72)
03-21 22:13:06.197: E/AndroidRuntime(25342): at com.example.androidspring.MainActivity$DownloadFilesTask.doInBackground(MainActivity.java:1)
03-21 22:13:06.197: E/AndroidRuntime(25342): at android.os.AsyncTask$2.call(AsyncTask.java:287)
03-21 22:13:06.197: E/AndroidRuntime(25342): at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:305)
03-21 22:13:06.197: E/AndroidRuntime(25342): ... 5 more
03-21 22:13:06.197: E/AndroidRuntime(25342): Caused by: org.codehaus.jackson.map.exc.UnrecognizedPropertyException: Unrecognized field "Loja" (Class com.example.androidspring.Loja), not marked as ignorable
03-21 22:13:06.197: E/AndroidRuntime(25342): at [Source: org.apache.http.conn.EofSensorInputStream@422e3c58; line: 1, column: 10] (through reference chain: com.example.androidspring.Loja["Loja"])
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.codehaus.jackson.map.deser.StdDeserializationContext.unknownFieldException(StdDeserializationContext.java:267)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.codehaus.jackson.map.deser.std.StdDeserializer.reportUnknownProperty(StdDeserializer.java:649)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.codehaus.jackson.map.deser.std.StdDeserializer.handleUnknownProperty(StdDeserializer.java:635)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.codehaus.jackson.map.deser.BeanDeserializer.handleUnknownProperty(BeanDeserializer.java:1355)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.codehaus.jackson.map.deser.BeanDeserializer.deserializeFromObject(BeanDeserializer.java:717)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.codehaus.jackson.map.deser.BeanDeserializer.deserialize(BeanDeserializer.java:580)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.codehaus.jackson.map.ObjectMapper._readMapAndClose(ObjectMapper.java:2723)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.codehaus.jackson.map.ObjectMapper.readValue(ObjectMapper.java:1914)
03-21 22:13:06.197: E/AndroidRuntime(25342): at org.springframework.http.converter.json.MappingJacksonHttpMessageConverter.readInternal(MappingJacksonHttpMessageConverter.java:122)
03-21 22:13:06.197: E/AndroidRuntime(25342): ... 15 more
据我所知，它不承认我的Loja课程。但它们完全一样！还有什么呢？
答案 0 :(得分：11)
异常似乎表明，虽然Jackson映射器试图从返回的json创建Loja，但它遇到了一个带有Loja键的json字段，它不知道如何处理（因为名为“Loja”的Loja java对象中没有字段，因此失败了。这似乎是合乎逻辑的，因为看起来服务器正在有效地返回这个json结构：
{"Loja":{"nome":"teste","xValue":"20","yValue":"30","andar":"1"}}
因此，您有两个主要选项。
getForObject()或对于第一个，您可以让您的客户getForObject()传递包含Object的{{1}}。
像这样的课：
Loja与此通话一起使用：
class MyObj {
private Loja Loja;
public void setLoja(Loja loja) {
this.Loja = loja;
}
public Loja getLoja() {
return this.Loja;
}
}
或者，您可以让服务器返回restTemplate.getForObject(url, MyObj.class);
对象字段作为模型的一部分，或者更好的是，返回您创建的Loja对象的实例。 Spring很聪明，可以利用Jackson将你的POJO变成你期望的正确的json模型。
Loja哪会产生这个json：
@RequestMapping("/android/played")
public Loja getLoja() {
System.out.println("Android Resquest.");
return new Loja("teste", "20", "30", "1");
}
客户端的{"nome":"teste","xValue":"20","yValue":"30","andar":"1"}
方法可以通过当前的方式读取。
答案 1 :(得分：2)
我发现我的代码出了什么问题。尼古拉斯是对的，我的JSON就像是：
{"Loja":{"nome":"teste","xValue":"20","yValue":"30","andar":"1"}}
并且在发送到Android应用程序时，Loja字段未被识别。
这就是我所做的：我创建了另一个名为Lojas的类，它只有一个Loja List。
public class Lojas {
private List<Loja> lojas;
public Lojas() {
}
public List<Loja> getLojas() {
return lojas;
}
public void setLojas(List<Loja> lojas) {
this.lojas = lojas;
}
}
在我的getLoja（）方法中，我将返回创建的这个全新方法：
public Lojas getLoja() {
RestTemplate restTemplate = new RestTemplate();
restTemplate.getMessageConverters().add(new MappingJacksonHttpMessageConverter());
restTemplate.setRequestFactory(new HttpComponentsClientHttpRequestFactory());
String url = BASE_URL + "/android/played.json";
return restTemplate.getForObject(url, Lojas.class);
}
这样，我可以立刻请求一个Loja对象列表。但要做到这一点，我不得不改变服务器端总是返回一个Loja对象列表：
@RequestMapping("/android/played")
public ModelAndView getLoja() {
System.out.println("Android Request.");
List<Loja> list = new ArrayList<Loja>();
list.add(new Loja("teste", "20", "30", "1"));
list.add(new Loja("teste2", "20", "30", "1"));
ModelAndView mav = new ModelAndView();
mav.addObject("lojas", list);
return mav;
}
答案 2 :(得分：2)
我使用JSON Post解决了这个RestTemplate问题。
HTTP_HEADERS.setContentType(MediaType.APPLICATION_JSON);
final ObjectWriter ow = new ObjectMapper().writer().withDefaultPrettyPrinter();
final String body = ow.writeValueAsString(convertedFireCsvBeans);
final HttpEntity<String> entity = new HttpEntity<>(body, HTTP_HEADERS);
final ParameterizedTypeReference<String> responseType = new ParameterizedTypeReference<String>(){};
final ResponseEntity<String> responseEntity = REST_TEMPLATE.exchange(url, HttpMethod.POST, entity, responseType);
if(responseEntity.getStatusCode() != null){
System.out.println("HTTP STATUS CODE: " + responseEntity.getStatusCode() +
"\n"+ responseEntity.getStatusCode().getReasonPhrase());
httpStatus = responseEntity.getStatusCode();
}
主要是我的HttpEntity我必须设置为String。
我希望它可以帮助别人。
答案 3 :(得分：0)
明确定义一个空的构造函数对我的情况有所帮助，否则我使用restTemplate的测试将失败。
答案 4 :(得分：0)
在父对象中不引入任何args构造函数将起作用。