Spring引导JPA - 没有带OneToMany关系的嵌套对象的JSON [英] Spring boot JPA - JSON without nested object with OneToMany relation

问题描述

我有一个处理对象的一些ORM映射的项目（有一些 @OneToMany 关系等等。）

我使用REST接口来处理这些对象，并使用Spring JPA在API中管理它们。

这是我的一个POJO示例：

  @Entity 
 public class Flight {
 
 @Id 
 @GeneratedValue（strategy = GenerationType.AUTO）
私人长ID; 
私人字符串名称; 
 private String dateOfDeparture; 
私人双倍距离; 
私人双重价格; 
 private int席位; 
 
 @ManyToOne（fetch = FetchType.EAGER）
 private Destination fromDestination; 
 
 @ManyToOne（fetch = FetchType.EAGER）
私人Destination toDestination; 
 
 @OneToMany（fetch = FetchType.EAGER，mappedBy =flight）
 private List< Reservation>保留; 
} 
  

提出请求时，我必须在JSON中指定所有内容：

  {
id：0，
reservations：[
 {} 
 $ bname：string，
dateOfDeparture：string，
距离：0，
price：0，
seat：0，
from：{
id：0，
name：string
}，
to ：{
id：0，
name：string
} 
} 
  

我更喜欢的是实际指定引用对象的id而不是它们的整体，如下所示：

<$ p
id：0，
reservations：[
{}
]，
name：字符串，
dateOfDeparture：字符串，
距离：0，
价格：0，
席位：0，
从：1，
到：2

甚至可能吗？有人能给我一些关于如何做到这一点的见解吗？我只找到如何做相反的教程（我已经有了解决方案）。 是的，这是可能的。

为此，您应该将一对Jackson注释用于实体模型：

  @JsonIdentityInfo（generator = ObjectIdGenerators.PropertyGenerator.class，property =id）
 @JsonIdentityReference（alwaysAsId = true）
 protected from; 
  

您的序列化JSON将会看起来不是这样：

  {
from：{
id：3，
description：New-York
} 
} 
  

像这样：

  {
from：3 
} 
  

正如官方文档：
$ b

@JsonIdentityReference - 可用于
的可选注释自定义引用的详细信息到Object
Identity已启用（请参阅 JsonIdentityInfo ）

alwaysAsId = true 用作标记来指示是否所有引用的
值都是b e序列化为ids（true）;

注意如果使用'true'的值，反序列化可能需要额外的上下文信息，使用自定义ID
解析器 - 默认处理可能不够。

I have a project which deals with some ORM mapping of objects (there are some @OneToMany relations etc).

I am using REST interface to treat these objects and Spring JPA to manage them in the API.

This is an example of one of my POJOs:

@Entity
public class Flight {

  @Id
  @GeneratedValue(strategy = GenerationType.AUTO)
  private long id;
  private String name;
  private String dateOfDeparture;
  private double distance;
  private double price;
  private int seats;

  @ManyToOne(fetch = FetchType.EAGER)
  private Destination fromDestination;

  @ManyToOne(fetch = FetchType.EAGER)
  private Destination toDestination;

  @OneToMany(fetch = FetchType.EAGER, mappedBy = "flight")
  private List<Reservation> reservations;
}

When making a request, I have to specify everything in the JSON:

{
  "id": 0,
  "reservations": [
    {}
  ],
  "name": "string",
  "dateOfDeparture": "string",
  "distance": 0,
  "price": 0,
  "seats": 0,
  "from": {
    "id": 0,
    "name": "string"
  },
  "to": {
    "id": 0,
    "name": "string"
  }
}

What I would prefer, is actually specifying the id of referenced object instead of their whole bodies, like this:

{
  "id": 0,
  "reservations": [
    {}
  ],
  "name": "string",
  "dateOfDeparture": "string",
  "distance": 0,
  "price": 0,
  "seats": 0,
  "from": 1,
  "to": 2
}

Is that even possible? Could someone give me some insight on how to do this? I am only finding tutorials on how to do the opposite (the solution I already have).

解决方案

Yes, it is possible.

For this purpose you should use pair of Jackson annotations to your entity model:

@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")
@JsonIdentityReference(alwaysAsId = true)
protected Location from;  

Your serialized JSON will look instead of this:

{
    "from": {
        "id": 3,
        "description": "New-York"
    } 
}

like this:

{
    "from": 3
}

As mentioned in official documentation:

@JsonIdentityReference - optional annotation that can be used for customizing details of a reference to Objects for which "Object Identity" is enabled (see JsonIdentityInfo)

alwaysAsId = true used as marker to indicate whether all referenced values are to be serialized as ids (true);

Note that if value of 'true' is used, deserialization may require additional contextual information, and possibly using a custom id resolver - the default handling may not be sufficient.

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