Question

当序列化具有其他对象引用的Java对象时，我只需要序列化嵌套对象的一个属性（通常是外键的情况，因此序列化对象引用的“id”属性）。其他一切。

例如，我有两个类需要序列化为JSON＆amp; XML（为清楚起见，删除了JPA注释）：

关系：用户 - ＆gt;（一对多）地址信息; 另外：AddressInformation - ＆gt;（一对一）用户

@XmlRootElement
public class User {
    private String id;
    private String firstName;
    private String lastName;
    private String email;
    private AddressInformation defaultAddress;
    private Set<AddressInformation> addressInformation;

    public User() {
    }

    @JsonProperty(value = "id")
    @XmlAttribute(name = "id")
    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    @JsonProperty(value = "firstname")
    @XmlAttribute(name = "firstname")
    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    @JsonProperty(value = "lastname")
    @XmlAttribute(name = "lastname")
    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    @JsonProperty(value = "email")
    @XmlAttribute(name = "email")
    public String getEmail() {
        return email;
    }

    public void setEmail(String email) {
        this.email = email;
    }

    @JsonIgnore
    public Set<AddressInformation> getAddressInformation() {
        return addressInformation;
    }

    public void setAddressInformation(Set<AddressInformation> addressInformation) {
        this.addressInformation = addressInformation;
    }

    @JsonProperty(value = "defaultaddress")
    @JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")
    public AddressInformation getDefaultAddress() {
        return defaultAddress;
    }

    public void setDefaultAddress(AddressInformation defaultAddress) {
        this.defaultAddress = defaultAddress;
    }
}

AddressInformation：

@XmlRootElement
public class AddressInformation  {
    private String id;
    private String address;
    private String details;
    private User user;

    @JsonProperty(value = "id")
    @XmlAttribute(name = "id")
    public String getId() {
        return id;
    }

    public void setId(String id) {
        this.id = id;
    }

    @JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")
    public User getUser() {
        return user;
    }

    public void setUser(User user) {
        this.user = user;
    }

    @JsonProperty(value = "details")
    @XmlAttribute(name = "details")
    public String getDetails() {
        return details;
    }

    public void setDetails(String details) {
        this.details = details;
    }

    @JsonProperty(value = "address")
    @XmlAttribute(name = "address")
    public String getAddress() {
        return address;
    }

    public void setAddress(String address) {
        this.address = address;
    }

    public AddressInformation() {
        super();
    }
}
enter code here

例如，在序列化用户时，我需要：

{
  "id" : "idofuser01",
  "email" : "some.email@gmail.com",
  "status" : "OK",
  "firstname" : "Filan",
  "lastname" : "Ovni",
  "defaultaddressid" : "idofaddress01",
}
enter code here

序列化AddressInformation时：

{
  "id" : "idofaddress01",
  "address" : "R.8. adn",
  "details" : "blah blah",
  "userid" : "idofuser01",
}

我试过@JsonManageReference＆amp; @JsonBackReference没有成功。如您所见，我也尝试了@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")

Answer 1

使用Jackson 2.1+找到了一种方法。

使用（仅选择id的{{1}}属性）注释对象引用：

AddressInformation

序列化效果很好。

Answer 2

您可以为此类实现自定义反序列化器，并在User类中使用它。示例实现：

class AddressInformationIdJsonSerializer extends JsonSerializer<AddressInformation> {
    @Override
    public void serialize(AddressInformation value, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException {
        jgen.writeString(value.getId());
    }
}

User类中的配置：

@JsonProperty(value = "defaultaddress")
@JsonSerialize(using = AddressInformationIdJsonSerializer.class)
public AddressInformation getDefaultAddress() {
    return defaultAddress;
}

###实现一个界面的所有类的通用解决方案###
您可以创建包含String getId()方法的接口：

interface Identifiable {
    String getId();
}

此界面的序列化程序可能如下所示：

class IdentifiableJsonSerializer extends JsonSerializer<Identifiable> {
    @Override
    public void serialize(Identifiable value, JsonGenerator jgen, SerializerProvider provider) throws IOException, JsonProcessingException {
        jgen.writeString(value.getId());
    }
}

现在，您可以将此序列化程序用于所有Identifiable实现。例如：

@JsonProperty(value = "defaultaddress")
@JsonSerialize(using = IdentifiableJsonSerializer.class)
public AddressInformation getDefaultAddress() {
    return defaultAddress;
}

当然：AddressInformation必须实现此接口：

class AddressInformation implements Identifiable {
    ....
}

杰克逊，序列化参考的一个属性

2 个答案: