将 json 对象转换为 java bean 类或 java pojo [重复]

Question

我有一个 json 对象

{ "Children": { "name": "rahul", "value": 1231, "calculated": false } }

并且需要转换成一个 java bean 类。但是在编写代码并打印 java 对象之后，它应该以 java 对象的形式打印整个 json 对象，但它只打印里面有 null 的 Children。它没有访问 Children 属性。

class Children {
    private String name;
    private int value;
    private boolean calculated;

    public Children(String name, int value, boolean calculated) {
        this.name = name;
        this.value = value;
        this.calculated = calculated;
    }

    public Children() {

    }

    // Add getters and setters for the fields...

    public String getName() {
        return name;
    }



    public int getValue() {
        return value;
    }



    public boolean isCalculated() {
        return calculated;
    }

    public void setName(String name) {
        this.name = name;
    }

    public void setValue(int value) {
        this.value = value;
    }

    public void setCalculated(boolean calculated) {
        this.calculated = calculated;
    }

    @Override
    public String toString() {
        return "Children{" +
                "name='" + name + '\'' +
                ", value=" + value +
                ", calculated=" + calculated +
                '}';
    }
}
class Itemcontainer extends Children
{

    Children children;
    public Children getChildren() {
        return children;
    }

    public void setChildren(Children children) {
        this.children = children;
    }

    @Override
    public String toString() {
        return "Itemcontainer{" +
                "children=" + children +
                '}';
    }
}


public class Main {
    public static void main(String[] args) throws IOException {



        ObjectMapper object = new ObjectMapper();
        object.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false);

        Itemcontainer itemcontainer;

        try {
           

            itemcontainer = object.readValue(new  File("E:\\json_home_impl\\saudip_3.json"),Itemcontainer.class );

            }
        catch (IOException e) {
            throw new RuntimeException(e);
        }

        System.out.println(itemcontainer.toString());
       

    }
}

得到输出为

{Children= null}

，正如我所期望的那样

 {"Children" { "name"= "rahul", "value"= 1231, "calculated"= false}}

Answer 1

存在大小写问题：在您的 json 中，您使用名称“Children”作为您的属性，但您的映射期望该字段被命名为“children”。由于您不会因未知属性而失败，因此它会被忽略。

选项：要么将 JSON 更改为具有小写的“children”属性，要么将 @JsonProperty 注释添加到您的映射中：

class Itemcontainer extends Children
{

    @JsonProperty("Children")
    Children children;
    public Children getChildren() {
        return children;
    }

    public void setChildren(Children children) {
        this.children = children;
    }

    @Override
    public String toString() {
        return "Itemcontainer{" +
                "children=" + children +
                '}';
    }
}