杰克逊：尝试将 JSON 反序列化为 Java 对象时未解析的前向引用

Question

我已经陷入了一个问题很长一段时间了，我渴望解决它。我有一个对象，其中包含带有ID相互引用的对象，我需要将其反序列化为对象，但是在尝试这样做时，我不断得到。JacksonJSONUnresolved forward references exception

我尝试过使用上面的注释类，但这没有产生任何结果。Jackson@JsonIdentityInfo

JSON 示例：

{

  "customer": [

    {

      "id": 1,

      "name": "Jane Gallow",

      "age": 16

    },

    {

      "id": 2,

      "name": "John Sharp",

      "age": 20

    },

  ],

  "shoppingcart": [

    {

      "id": 1,

      "customer": 2,

      "orderDate": "2015-02-19"

    }

  ]

}

客户类

@JsonIdentityInfo(generator= ObjectIdGenerators.PropertyGenerator.class, property="id",scope = Customer.class)

@JsonIdentityReference(alwaysAsId = true)

public class Customer {

    private int id = -1;

    private String name;

    private int age;

    //getters, setters

}

购物车类

<!-- language: java -->

@JsonIdentityInfo(generator= ObjectIdGenerators.PropertyGenerator.class, property="id",scope = ShoppingCart.class)

public class ShoppingCart {

    private int id = -1;

    private Customer customer;

    @JsonDeserialize(using = LocalDateDeserializer.class)

    @JsonSerialize(using = LocalDateSerializer.class)

    private LocalDate orderDate = LocalDate.now();

    //getters, setters

}

我希望给我一个对象，该对象具有对id中的对象的引用（在本例中为约翰·夏普）。但是我只是无法让它工作，当我尝试使用方法从JSON读取时，它会给我。JacksonShoppinCartCustomer2"main" com.fasterxml.jackson.databind.deser.UnresolvedForwardReferenceObjectMapper.readValue()

Answer 1

您应该对属性使用&nbsp;JsonIdentity 引用。@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id", scope = ShoppingCart.class)class ShoppingCart {&nbsp; &nbsp; private int id = -1;&nbsp; &nbsp; @JsonIdentityReference&nbsp; &nbsp; private Customer customer;&nbsp; &nbsp; private LocalDate orderDate;&nbsp; &nbsp; // getters, setters, toString}@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id", scope = Customer.class)class Customer {&nbsp; &nbsp; private int id = -1;&nbsp; &nbsp; private String name;&nbsp; &nbsp; private int age;&nbsp; &nbsp; // getters, setters, toString}简单示例：import com.fasterxml.jackson.annotation.JsonIdentityInfo;import com.fasterxml.jackson.annotation.JsonIdentityReference;import com.fasterxml.jackson.annotation.JsonProperty;import com.fasterxml.jackson.annotation.ObjectIdGenerators;import com.fasterxml.jackson.databind.ObjectMapper;import com.fasterxml.jackson.databind.SerializationFeature;import com.fasterxml.jackson.datatype.jsr310.JavaTimeModule;import java.io.File;import java.time.LocalDate;import java.util.List;public class JsonApp {&nbsp; &nbsp; public static void main(String[] args) throws Exception {&nbsp; &nbsp; &nbsp; &nbsp; File jsonFile = new File("./resource/test.json").getAbsoluteFile();&nbsp; &nbsp; &nbsp; &nbsp; ObjectMapper mapper = new ObjectMapper();&nbsp; &nbsp; &nbsp; &nbsp; mapper.registerModule(new JavaTimeModule());&nbsp; &nbsp; &nbsp; &nbsp; mapper.enable(SerializationFeature.INDENT_OUTPUT);&nbsp; &nbsp; &nbsp; &nbsp; System.out.println(mapper.readValue(jsonFile, Pojo.class));&nbsp; &nbsp; }}对于您的打印件：JSONPojo{customers=[Customer{id=1, name='Jane Gallow', age=16}, Customer{id=2, name='John Sharp', age=20}], shoppingCarts=[ShoppingCart{id=1, customer=Customer{id=2, name='John Sharp', age=20}, orderDate=2015-02-19}]}

Answer 2

在反序列化过程中，我对未解析的正向引用也有同样的问题。我使用 https://www.logicbig.com/tutorials/misc/jackson/json-identity-reference.html 上提供的示例来检查并解决我的问题。有一个非常简单的类：public class Order {&nbsp; private int orderId;&nbsp; private List<Integer> itemIds;&nbsp; private Customer customer;....}我向 Customer 类添加了@JsonCreator静态工厂方法，它没有导致任何问题。现在，当我将任何类型的@JsonCreator（静态方法或构造函数）添加到 Order 类时，我会立即获得未解析的前向引用。有趣的是，当我添加以下构造函数时，只有@JsonPropery个注释：&nbsp;public Order(@JsonProperty("orderId") int orderId, @JsonProperty("itemIds") List<Integer> itemIds, @JsonProperty("customer") Customer customer) {&nbsp; &nbsp; this.orderId = orderId;&nbsp; &nbsp; this.itemIds = itemIds;&nbsp; &nbsp; this.customer = customer;}它还会导致未解析的前向引用。要解决此问题，对于 Order 类，您必须创建无参数构造函数（可以是私有的）。否则，您将获得：线程“主”中的异常：无法构造实例（没有创建者，如默认构造存在）：无法从对象值反序列化（没有基于委托或基于属性的创建者）com.logicbig.example.Order并且不得有任何@JsonCreator带注释的方法和构造函数，或者具有@JsonProperty批注的构造函数。仅此而已。