JsonMappingException：无法实例化没有单长参数构造函数/工厂方法类型的值

Question

您好，尝试在 zk 框架上将 json 响应解析为 java 时出错

这是杰森的回复

{"currentTime":1355390722038,"text":"OK","data":{"limitExceeded":false,"references":{"stops":[],"situations":[],"trips":[],"routes":[],"agencies":[{"id":"AG1","privateService":false,"phone":"","timezone":"Asia/Kuala_Lumpur","disclaimer":"","name":"Panorama","lang":"en","url":"http://www.allcompanyonline.com/company/33/8/26/62438/PM-Cultural--Tourism-Sdn-Bhd.html"}]},"list":[{"lonSpan":0.1766824722290039,"lon":102.2011971473685,"agencyId":"AG1","lat":2.2659808772471948,"latSpan":0.15555363245723042}]},"code":200,"version":2}

我一直收到这个错误

com.fasterxml.jackson.databind.JsonMappingException: Can not instantiate value of type [simple type, class com.panorama.tripplan.pojo.Agency_coverage] from Long integral number; no single-long-arg constructor/factory method

这里是根变量的 pojo

@JsonRootName(value = "currentTime")
public class Agency_coverage{
private Number code;
private Number currentTime;
private Data data;
private String text;
private Number version;

public Agency_coverage(){}

public Number getCode(){
    return this.code;
}
public void setCode(Number code){
    this.code = code;
}
public Number getCurrentTime(){
    return this.currentTime;
}
public void setCurrentTime(Number currentTime){
    this.currentTime = currentTime;
}
public Data getData(){
    return this.data;
}
public void setData(Data data){
    this.data = data;
}
public String getText(){
    return this.text;
}
public void setText(String text){
    this.text = text;
}
public Number getVersion(){
    return this.version;
}
public void setVersion(Number version){
    this.version = version;
}

这是我尝试调用 api 并解析响应的代码

java.net.URLConnection connection = new URL(url + "?" + query+ "&" + queryLat+ "&" + queryLon).openConnection();
       connection.setRequestProperty("Accept-Charset", charset);
       if ( connection instanceof HttpURLConnection)
       {
         //Jackson 2.X configuration settings
            mapper.configure(DeserializationFeature.UNWRAP_ROOT_VALUE, true);

            mapper.configure(DeserializationFeature.ACCEPT_SINGLE_VALUE_AS_ARRAY, true);
            mapper.configure(DeserializationFeature.ACCEPT_EMPTY_STRING_AS_NULL_OBJECT, true);
            mapper.configure(DeserializationFeature.USE_JAVA_ARRAY_FOR_JSON_ARRAY, true);
            mapper.configure(DeserializationFeature.READ_ENUMS_USING_TO_STRING, true);
            mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

            //Tell Jackson to expect the JSON in PascalCase, instead of camelCase
            mapper.setPropertyNamingStrategy(new PropertyNamingStrategy.PascalCaseStrategy());

          HttpURLConnection httpConnection =(HttpURLConnection) connection;  
        //Make the HTTP request, and deserialize the JSON response into the object

          Agency_coverage agencyData= new Agency_coverage();
          agencyData = mapper.readValue(agenturl, Agency_coverage.class);


          //System.out.println(agencyData.getData().getList().get(0).);
          System.out.println(agencyData.getCurrentTime());
          System.out.println(httpConnection.getResponseCode());

       }

如果你能理解这个问题，请帮忙，我对java开发有点新，我全力以赴

Answer 1

第一件事是删除@aymeric 建议的@JsonRootName(value = "currentTime") 注释。此注释告诉 Jackson 映射器，当您的 POJO 将此事物称为“Agency_coverage”时，匹配的 JSON 对象将称为“currentTime”。事实并非如此。

现在，除此之外，您的 JSON 应该看起来更像这样：

{"Agency_coverage":{"currentTime":1355390722038,"text":"OK","data":"etc..."}}

这里的不同之处在于您提供了让 Jackson 展开的根元素。

您现在可以看到 Jackson 的错误在说什么。使用您提供的 JSON 和您使用的注释，它发现 currentTime 作为映射到 Agency_coverage POJO 的 JSON 元素，但 currentTime 的值是 long 并且无法映射。