使用外部API并將JSON字串轉換為bean不會初始化屬性。NULL物件屬性。春季啟動，Maven

這是我從 API 得到的回應。

{"get":"statistics","parameters":{"country":"romania"},"errors":[],"results":1,"response":[{"continent":"Europe", “國家”：“羅馬尼亞”，“人口”：19016885，“病例”：{“新”：“ 4521”，“活躍”：156487，“關鍵”：431，“恢復”：2606660，“1M_pop”： "148707","total":2827936},"deaths":{"new":" 35","1M_pop":"3407","total":64789},"tests":{"1M_pop":" 1149381","總計":21857638},"天":"2022-03-24","時間":"2022-03-24T07:30:04 00:00"}]}

@RestController
public class CovidTrackerRestController {
    
    @GetMapping("/hello")
    public String showCovidInformation() {
        
        // connect to a covid database
                        
        HttpRequest request = HttpRequest.newBuilder()
                .uri(URI.create("https://covid-193.p.rapidapi.com/statistics?country=romania"))
                .header("X-RapidAPI-Host", "covid-193.p.rapidapi.com")
                .header("X-RapidAPI-Key", "mykey")
                .method("GET", HttpRequest.BodyPublishers.noBody())
                .build();
        HttpResponse<String> response = null;
        
        try {
            response = HttpClient.newHttpClient().send(request, HttpResponse.BodyHandlers.ofString());
        } catch (IOException | InterruptedException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
                        
        // get the information
        String responseString = response.body();
        System.out.println(responseString);
        
        Response romaniaData = null;
        
        try {
            romaniaData = new ObjectMapper().configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false)
                            .readValue(responseString, Response.class);
        } catch (JsonProcessingException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
            
        
        // format the information
        System.out.println(romaniaData);
                
        
        // send the information to html page
        return "/tracker";
    }
}

And this is my Bean class which is annotated with @Bean in the configurator class alonside the RestTemplate bean. Other properties such as Cases, Deaths etc are configured same as Response class except being declared as @Bean in the configurator because from what I know once I declare a class @Bean then other references contained automatically become beans as well.

@JsonIgnoreProperties(ignoreUnknown = true)
public class Response {

    @JsonProperty("country")
    private String country;
    
    @JsonProperty("cases")
    private Cases cases;
    
    @JsonProperty("deaths")
    private Deaths deaths;
    
    @JsonProperty("day")
    private String day;
    
    @JsonProperty("time")
    private String time;
    
    @JsonProperty("test")
    private Tests tests;
    
    public String getCountry() {
        return country;
    }
    public void setCountry(String country) {
        this.country = country;
    }

uj5u.com熱心網友回復：

是的，你的班級應該是這樣的：

public class ResponseWrapper {
  public List<Response> parameters;

  public setParameters(List<Response> parameters) {
    this.parameters = parameters;
  }
  public List<Response> getParameters() {
    return parameters;
  }
}

當您發布課程時，課程Response就是您的課程。您的類必須具有與 JSON 相同的結構

uj5u.com熱心網友回復：

您的 java 類需要是接收到的 json 的精確表示。讓我們稱之為Wrapper：

public class Wrapper {

    @JsonProperty("response")
    private List<Response> responses;

    public List<Response> getResponses() {
        return this.responses;
    }

    public void setResponses(List<Response> responses) {
        this.responses = responses;
    }

    @Override
    public String toString() {
        return "Wrapper{"  
                "responses="   responses  
                '}';
    }
}

我省略了一些屬性 - get,results等。看起來你不需要它們。然后反序列化將如下所示：

Wrapper data = null;
try {
    data = new ObjectMapper()
            .configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false)
            .readValue("json", Wrapper.class);
} catch (IOException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}
System.out.println(data);

幾點注意事項：

1. 如果 json 屬性名與類中的欄位名匹配，則不需要@JsonProperty
2. 對于測驗欄位注釋應該是 - @JsonProperty("tests")。屬性是tests,不是test

如果你真的想扔掉剩下的資料，并且只需要response屬性，那么你需要撰寫自定義反序列化器并處理 json 樹。例如，您可以在我的答案或本指南中看到如何做到這一點。像這樣，您可以將回應 json 決議到您的類，即使它們的結構不匹配。

標籤：java json spring spring-boot maven

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