Java Fasterxml-如何从json文件中排除对象?
我将以Java Fasterxml-如何从json文件中排除对象?,java,json,hibernate,jackson,fasterxml,Java,Json,Hibernate,Jackson,Fasterxml,我将以JSON格式发送的实体映射到服务。这是我的实体 @Entity @Table(name = "company") @JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id") public class Company implements Serializable { @Id @GeneratedValue(strategy = GenerationTy
JSON
格式发送的实体映射到服务。这是我的实体
@Entity
@Table(name = "company")
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")
public class Company implements Serializable {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private Integer id;
@Column
private String name;
@OneToMany(mappedBy = "company")
@Cascade(value = CascadeType.ALL)
private Collection<Employee> employees;
但我得到的不是合适的json格式
{
"id": 1,
"name": "Tim",
"company": {
"id": 1,
"name": "Microsoft",
"employees": [1, {
"id": 5,
"name": "Jack",
"company": 1
}, {
"id": 6,
"name": "Jack",
"company": 1
}, {
"id": 7,
"name": "Jack",
"company": 1
}, {
"id": 8,
"name": "Tommy",
"company": 1
}]
}
}
但就像我说的,我不需要“公司”中的“员工”对象。如何在我的JSON
文件中排除它?您可以使用它将“employees”对象从“company”中排除。下面是一个基于代码的示例,演示了该方法:
public class JacksonReferences {
@JsonIdentityInfo(generator = ObjectIdGenerators.None.class, property = "id")
static public class Company {
public Integer id;
public String name;
@JsonManagedReference
public Collection<Employee> employees;
public Company(Integer id, String name, Collection<Employee> employees) {
this.id = id;
this.name = name;
this.employees = employees;
}
}
@JsonIdentityInfo(generator = ObjectIdGenerators.None.class, property = "id")
static public class Employee {
public Integer id;
public String name;
@JsonBackReference
public Company company;
public Employee(Integer id, String name, Company company) {
this.id = id;
this.name = name;
this.company = company;
}
}
public static void main(String[] args) throws IOException {
Company company1 = new Company(1, "Microsoft", new ArrayList<Employee>());
Company company2 = new Company(2, "Google", new ArrayList<Employee>());
Employee employee1 = new Employee(1, "John", company1);
company1.employees.add(employee1);
Employee employee2 = new Employee(2, "Tim", company1);
company1.employees.add(employee2);
Employee employee3 = new Employee(3, "Bob", company2);
company2.employees.add(employee3);
ObjectMapper mapper = new ObjectMapper();
System.out.println("JSON for company #1:");
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(company1));
System.out.println("JSON for employee #1:");
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(employee1));
System.out.println("JSON for company #2:");
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(company2));
}
}
您可以尝试使用
@JsonIgnore
注释。它应该有助于解决这个问题。相同的输出没有@JsonIdentityInfo。
public class JacksonReferences {
@JsonIdentityInfo(generator = ObjectIdGenerators.None.class, property = "id")
static public class Company {
public Integer id;
public String name;
@JsonManagedReference
public Collection<Employee> employees;
public Company(Integer id, String name, Collection<Employee> employees) {
this.id = id;
this.name = name;
this.employees = employees;
}
}
@JsonIdentityInfo(generator = ObjectIdGenerators.None.class, property = "id")
static public class Employee {
public Integer id;
public String name;
@JsonBackReference
public Company company;
public Employee(Integer id, String name, Company company) {
this.id = id;
this.name = name;
this.company = company;
}
}
public static void main(String[] args) throws IOException {
Company company1 = new Company(1, "Microsoft", new ArrayList<Employee>());
Company company2 = new Company(2, "Google", new ArrayList<Employee>());
Employee employee1 = new Employee(1, "John", company1);
company1.employees.add(employee1);
Employee employee2 = new Employee(2, "Tim", company1);
company1.employees.add(employee2);
Employee employee3 = new Employee(3, "Bob", company2);
company2.employees.add(employee3);
ObjectMapper mapper = new ObjectMapper();
System.out.println("JSON for company #1:");
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(company1));
System.out.println("JSON for employee #1:");
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(employee1));
System.out.println("JSON for company #2:");
System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(company2));
}
}
JSON for company #1:
{
"id" : 1,
"name" : "Microsoft",
"employees" : [ {
"id" : 1,
"name" : "John"
}, {
"id" : 2,
"name" : "Tim"
} ]
}
JSON for employee #1:
{
"id" : 1,
"name" : "John"
}
JSON for company #2:
{
"id" : 2,
"name" : "Google",
"employees" : [ {
"id" : 3,
"name" : "Bob"
} ]
}