JSON使用hibernate jpa序列化和反序列化以在JSON响应中将父对象变为子对象 [英] JSON serializing and deserializing with hibernate jpa to have parent object into child in JSON response
问题描述
我正在用Spring框架，Hibernate和JSON开发其他Web应用程序。请假设我有两个实体，如下所示：
BaseEntity.java
@MappedSuperclass
@JsonIdentityInfo（generator = ObjectIdGenerators.PropertyGenerator.class，property =id）
公共抽象类BaseEntity实现Serializable {
@Id
@GeneratedValue（strategy = GenerationType.IDENTITY）
私人长ID;
public long getId（）{
return id;
}
}
University.java
public class University扩展BaseEntity {
private String uniName;
@OneToMany（cascade = CascadeType.ALL，fetch = FetchType.EAGER，orphanRemoval = true）
@JoinColumn（name =university_id）
private List< Student> students = new ArrayList<>（）;
// setter a getter
}
Student.java
public class Student扩展BaseEntity {
private String stuName;
@ManyToOne（fetch = FetchType.EAGER）
@JoinColumn（name =university_id，updatable = false，insertable = false）
私立大学;
// setter a getter
}
我的休息API列表大学的每一件事情都正常工作，但我打电话给我的休息api时急切地列出学生我的JSON响应是
<$ c
$b
id：1，
stuName：st1，
university：{
id：1，
uniName：uni1
}
}，
{
id：2，
stuName：st2，
university：1
}
]
但我的< $ b
[
{
id：1，
stutName：st1，
university：
{
id：1，
uniName：uni1
}
}，
{
id：2，
stutName：st2，
university：
{
id：1，
uniName：uni 1
}
更新1 ：我的hibernate注释工作正常我有JSON问题
要求：
-
我需要热切地接受双方的提取（大学方面没问题）
-
每个学生都需要学生方面的大学对象（当我热切地提取学生时）
-
什么样的序列化或JSON配置我需要这样做来匹配我想要的响应？
更新2：如下所示：
$ b
@ManyToOne（fetch = FetchType.EAGER）
@JoinColumn（名称=university_id，updatable = false，可插入= false）
@JsonIgnoreProperties（value =students，allowSetters = true）
私立大学;
json响应仍然相同
我需要我的< 所需的响应是上面提到的。
谢谢
我有同样的问题。休眠（或eclipselink）不是问题。
JPA中唯一的约束是FetchType.EAGER 。
在BaseEntity中，我添加了一个标准方法
public String getLabel（）{
returnid：+ this.getId（）;
}
这种方法很抽象，但我有很多课， '
在父实体中，在这种情况下，大学，重写方法
@Override
public String getLabel {
return this.uniName;
$ / code> 对于每个父类，使用一个特定的字段作为实体的标签
定义MyStandardSerializer：
public class StandardJsonSerializer扩展JsonSerializer< EntityInterface> {
$ b @Override
public void serializeWithType（EntityInterface value，JsonGenerator jgen，SerializerProvider provider，TypeSerializer typeSer）throws IOException，JsonProcessingException {
serialize（value，jgen，provider）;
$ b @Override
public void serialize（EntityInterface value，JsonGenerator jgen，SerializerProvider provider）
抛出IOException，JsonProcessingException {
jgen.writeStartObject（） ;
jgen.writeNumberField（id，（long）value.getId（））;
jgen.writeStringField（label，value.getLabel（））;
jgen.writeEndObject（）;
}
}
在学生类中，在univrsity上添加：
$ $ p $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $ @ $ @Coin（name =university_id，updatable = false，insertable = false）
@JsonSerialize（使用= StandardJsonSerializer.class）
私立大学;
现在您已经解决了循环。
当您需要标签时，请覆盖父实体中的方法。
当您需要某些特定字段时，请创建一个特定的序列化程序。
I am developing rest web app with spring framework, Hibernate and JSON. Please Assume that I have two entities like below:
BaseEntity.java
@MappedSuperclass
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class,property = "id" )
public abstract class BaseEntity implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;
public long getId() {
return id;
}
}
University.java
public class University extends BaseEntity {
private String uniName;
@OneToMany(cascade = CascadeType.ALL, fetch = FetchType.EAGER,orphanRemoval = true)
@JoinColumn(name = "university_id")
private List<Student> students=new ArrayList<>();
// setter an getter
}
Student.java
public class Student extends BaseEntity{
private String stuName;
@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "university_id",updatable = false,insertable = false)
private University university;
// setter an getter
}
when I call my rest api to list University every things work fine as I expect, but when I call my rest api to list student eagerly my JSON response is
[
{
"id": 1,
"stuName": "st1",
"university": {
"id": 1,
"uniName": "uni1"
}
},
{
"id": 2,
"stuName": "st2",
"university": 1
}
]
but my desired response is:
[
{
"id": 1,
"stutName": "st1",
"university":
{
"id": 1,
"uniName": "uni1"
}
},
{
"id": 2,
"stutName": "st2",
"university":
{
"id": 1,
"uniName": "uni1"
}
}
Update 1: my hibernate annotation working fine I have JSON issue
Requirements :
I need both side fetch eagerly(the university side is Ok)
I need university object in student side for every student(when I fetching student eagerly)
What kind of serialization or JSON config I need to do that for matching my desired response?
Update 2:
by removing @JsonIdentityInfo and editing student side like below:
@ManyToOne(fetch = FetchType.EAGER) @JoinColumn(name = "university_id",updatable = false,insertable = false) @JsonIgnoreProperties(value = "students", allowSetters = true) private University university;
the json response still same I need my desired response that is mentioned above.
thanks
I had the same problem. Hibernate (or eclipselink) are not the problem. The only constraint in JPA is the FetchType.EAGER .
In the BaseEntity I have added a standard method
public String getLabel(){
return "id:"+this.getId();
}
this method would be abstract, but I had a lot of class and i didn't want to change it all so I added a default value.
In parent entity, in this case University, override the method
@Override
public String getLabel{
return this.uniName;
}
For each parent class, use a particular field as a label for your entity
Define a MyStandardSerializer:
public class StandardJsonSerializer extends JsonSerializer<EntityInterface> {
@Override
public void serializeWithType(EntityInterface value, JsonGenerator jgen, SerializerProvider provider, TypeSerializer typeSer) throws IOException, JsonProcessingException {
serialize(value, jgen, provider);
}
@Override
public void serialize(EntityInterface value, JsonGenerator jgen, SerializerProvider provider)
throws IOException, JsonProcessingException {
jgen.writeStartObject();
jgen.writeNumberField("id", (long) value.getId());
jgen.writeStringField("label", value.getLabel());
jgen.writeEndObject();
}
}
In the student class, on univrsity add:
@ManyToOne(fetch = FetchType.EAGER)
@JoinColumn(name = "university_id",updatable = false,insertable = false)
@JsonSerialize(using=StandardJsonSerializer.class)
private University university;
Now you have resolved circularity. When you need a label, override the method in the parent entity. When you need some particular fields, create a specific Serializer.
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