杰克逊,序列化参考的一个属性

Question

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,我需要序列化为JSON和XML(为清楚起见,删除了JPA注释):

关系:用户 - >(一对多)地址信息; 另外:AddressInformation - >(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;
    }
}

地址信息:

@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",
}

我曾尝试@JsonManageReference与@JsonBackReference没有成功.如你所见,我也试过了 @JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")

Answer 1

刚刚找到了使用Jackson 2.1+的方法.

使用(这将仅选择id属性AddressInformation)注释对象引用:

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

序列化工作得很好.