尝试将 JSON 字符串转换为 JavaPOJO 时遇到奇怪的问题

Question

Maven 依赖 - Eclipse IDE

<!-- https://mvnrepository.com/artifact/com.fasterxml.jackson.core/jackson-databind -->
        <dependency>
                <groupId>com.fasterxml.jackson.core</groupId>
                <artifactId>jackson-databind</artifactId>
                <version>2.9.8</version>
        </dependency>

没有任何注释，我得到

控制台错误

com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException：无法识别的字段 "id" (class com.optimight.prakash.general.Employee)，未标记为可忽略（0 个已知属性：]）在[来源：（字符串）”{"id":1,"name":"Lokesh Gupta","age":34,"location":"India"}” ;行：1，列：8]（通过引用链：com.optimight.prakash.general.Employee["id"]）在 com.fasterxml.jackson.databind.exc.UnrecognizedPropertyException.from(UnrecognizedPropertyException.java:61) 在 com.fasterxml.jackson.databind.DeserializationContext.handleUnknownProperty(DeserializationContext.java:823) 在 com.fasterxml.jackson.databind.deser.std.StdDeserializer.handleUnknownProperty(StdDeserializer.java:1153) 在 com.fasterxml.jackson.databind.deser.BeanDeserializerBase.handleUnknownProperty(BeanDeserializerBase.java:1589) 在 com.fasterxml.jackson.databind.deser.BeanDeserializerBase.handleUnknownVanilla(BeanDeserializerBase.java:1567) 在 com.fasterxml.jackson.databind.deser.BeanDeserializer.vanillaDeserialize(BeanDeserializer.java:294) 在 com.fasterxml.jackson.databind.deser.BeanDeserializer.deserialize(BeanDeserializer.java:151) 在 com.fasterxml.jackson.databind.ObjectMapper._readMapAndClose(ObjectMapper.java:4013) 在 com.fasterxml.jackson.databind.ObjectMapper.readValue(ObjectMapper.java:3004) 在 com.optimight.prakash.general.JSONtoJavaPOJO.main(JSONtoJavaPOJO.java:18)

带注释：@JsonIgnoreProperties(ignoreUnknown = true)

在 Employe Class 中，然后我得到

员工[id=null, name=null, age=null, location=null]

应用程序Class - 主要方法

    import java.io.IOException;
    import com.fasterxml.jackson.core.JsonGenerationException;
    import com.fasterxml.jackson.databind.JsonMappingException;
    import com.fasterxml.jackson.databind.ObjectMapper;

public class JSONtoJavaPOJO {
public static void main(String[] args) {
    String json = "{\"id\":1,\"name\":\"Lokesh Gupta\",\"age\":34,\"location\":\"India\"}";

    ObjectMapper mapper = new ObjectMapper();
    try
    {
        Employee emp = mapper.readValue(json, Employee.class);

        System.out.println(emp);
    }
    catch (JsonGenerationException e)
    {
        e.printStackTrace();
    } catch (JsonMappingException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
}

}

员工Class

import com.fasterxml.jackson.annotation.JsonIgnoreProperties;



//@JsonIgnoreProperties(ignoreUnknown = true)
public class Employee
{
    private Integer id;
    private String name;
    private Integer age;
    private String location;



    public Employee() {
    }


    public Employee(Integer id, String name, Integer age, String location) {
        super();
        this.id = id;
        this.name = name;
        this.age = age;
        this.location = location;
    }



    @Override
    public String toString() {
        return "Employee [id=" + id + ", name=" + name + ", age=" + age
                + ", location=" + location + "]";
    }
}

Answer 1

为您的员工添加 setter 和 getter class。

Answer 2

您可以使用项目 lombok 添加 setter 和 getter。最终员工 class 会看起来像这样。

@Data
@NoArgsConstructor
@AllArgsConstructor
@ToString
public class Employee
{
    private Integer id;
    private String name;
    private Integer age;
    private String location;
}

尝试将 JSON 字符串转换为 JavaPOJO 时遇到奇怪的问题

Facing weird problem while trying to convert JSON string into JavaPOJO

java

json

pojo