android - HttpMessageNotReadableException：无法读取 JSON：使用 Spring for Android 无法识别的字段

Question

我正在开发与我的服务器通信的 Android 应用程序。这种通信是通过 Spring 框架和 Jackson 完成的。我正在成功发送对我的服务器的请求，但没有收到响应。这是我所做的：

安卓应用：

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 类（我有两个版本。一个在 Android 应用程序中，另一个在服务器中。它们完全一样）：

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;
    }

有了这一切，我在 Android 应用程序中遇到了以下异常：

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

根据我的理解，它不承认我的 Loja 课程。但它们完全一样！还能是什么？

score 11 · Accepted Answer

该异常似乎表明，当 Jackson 映射器尝试Loja从返回的 json 创建一个时，它遇到了一个带有 Loja 键的 json 字段，它不知道如何处理（因为Lojajava 对象中没有字段命名为“Loja”），所以它失败了。这似乎是合乎逻辑的，因为服务器似乎有效地返回了这个 json 结构：

{"Loja":{"nome":"teste","xValue":"20","yValue":"30","andar":"1"}}

因此，您有两个主要选择。

更改传递给getForObject()或的对象
更改从服务器返回的 json。

对于第一个，您可以让您的客户getForObject()传递一个Object包含Loja.

像这个类的东西：

class MyObj {
    private Loja Loja;

    public void setLoja(Loja loja) {
        this.Loja = loja;
    }

    public Loja getLoja() {
        return this.Loja;
    }
}

与此调用一起使用：

restTemplate.getForObject(url, MyObj.class);

或者，您可以让您的服务器将Loja对象字段作为模型的一部分返回，或者更好的是，返回Loja您创建的对象的实例。Spring 足够聪明，可以利用 Jackson 将您的 POJO 转换为您期望的正确 json 模型。

@RequestMapping("/android/played")
public Loja getLoja() {
    System.out.println("Android Resquest.");

    return new Loja("teste", "20", "30", "1");
}

这会产生这个json：

{"nome":"teste","xValue":"20","yValue":"30","andar":"1"}

getForObject()您在客户端的方法可以按照当前的方式读取它。

score 2 · Accepted Answer

我发现我的代码有什么问题。Nicholas 是对的，我的 JSON 类似于：

{"Loja":{"nome":"teste","xValue":"20","yValue":"30","andar":"1"}}

并且字段 Loja 在发送到 Android 应用程序时未被识别。

这就是我所做的：我创建了另一个名为 Lojas 的类，它只不过是一个 Loja 列表。

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，我返回了这个创建的全新方法：

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，我可以立即请求 Loja 对象列表。但要做到这一点，我必须将服务器端更改为始终返回 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;
    }

score 2 · Accepted Answer

我解决了这个问题，JSON Post使用RestTemplate.

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();                    
                }

最主要的是HttpEntity我必须设置为String.

我希望它可以帮助某人。

score 0 · Accepted Answer

明确定义一个空的构造函数对我有帮助，否则我使用 restTemplate 的测试会失败。

score 0 · Accepted Answer

0

在父对象中不引入 args 构造函数将起作用。

于 2020-08-28T14:08:31.820 回答

android - HttpMessageNotReadableException：无法读取 JSON：使用 Spring for Android 无法识别的字段

5 回答 5

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