Fasterxml - How to exclude an object from json file?

Question

I have mapped entities which I send in JSON format to the service. Here is my entities

@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;

My Employee class

@Entity
@Table(name = "employee")
@JsonIdentityInfo(generator = ObjectIdGenerators.PropertyGenerator.class, property = "id")
public class Employee implements Serializable {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    Integer id;

    @Column
    String name;

    @ManyToOne()
    @Cascade(value = org.hibernate.annotations.CascadeType.ALL)
    @JoinColumn(name = "company_id", referencedColumnName = "id")
    private Company company;

But I'm getting not appropriate json format.

{
    "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
                        }]
    }
}

But like I said I don't need "employees" object in "company". How to exclude it in my JSON file?

You can try with `@JsonIgnore` annotation. It should help with this problem. — Michał Ziober, Apr 15 '14 at 17:53

score 1 · Accepted Answer · answered Apr 19 '14 at 09:21

You can use the Jackson's bi-directional references to exclude "employees" object from "company". Here is an example based on your code that demonstrate the approach:

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));
    }
}

The output is:

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"
  } ]
}

Same output without @JsonIdentityInfo. – Rajendra Thorat Mar 17 '15 at 16:52 — Rajendra Thorat, Mar 17 '15 at 16:52

Fasterxml - How to exclude an object from json file?

1 Answers1