java - 如何使用 jackson mixins 将 json 映射到具有不同结构的 java 对象

Question

如何转换这个json

{
    "name": "abc",
    "city": "xyz"
}

使用 Jackson mixin 到员工对象

//3rd party class//
public class Employee {
    public String name;
    public Address address;
}

//3rd party class//
public class Address {
    public String city;
}

Answer 1

通常，您会使用 @JsonUnwrapped 注释 address 字段，以便在序列化时解包(并在反序列化时包装)。但是因为你不能改变你的第 3 方类，所以你必须在 mixin 上做这个:

// Mixin for class Employee
abstract class EmployeeMixin {
    @JsonUnwrapped public Address address;
}

然后，创建一个包含所有特定“扩展”的模块。这可以通过子类化 Module 或 SimpleModule 来完成，或者通过如下组合 SimpleModule 来完成:

SimpleModule module = new SimpleModule("Employee");
module.setMixInAnnotation(Employee.class, EmployeeMixin.class);

第三，使用您的 ObjectMapper 注册模块:

ObjectMapper mapper = new ObjectMapper();
mapper.registerModule(module);

最后，享受序列化/反序列化的乐趣吧!

---

SimpleModule 子类的自包含完整示例:

public class TestJacksonMixin {

    /* 3rd party */
    public static class Employee {
        public String name;
        public Address address;
    }

    /* 3rd party */
    public static class Address {
        public String city;
    }

    /* Jackon Module for Employee */
    public static class EmployeeModule extends SimpleModule {
        abstract class EmployeeMixin {
            @JsonUnwrapped
            public Address address;
        }

        public EmployeeModule() {
            super("Employee");
        }

        @Override
        public void setupModule(SetupContext context) {
            setMixInAnnotation(Employee.class, EmployeeMixin.class);
        }
    }

    public static void main(String[] args) throws JsonProcessingException {
        Employee emp = new Employee();
        emp.name = "Bob";
        emp.address = new Address();
        emp.address.city = "New York";

        ObjectMapper mapper = new ObjectMapper();
        mapper.registerModule(new EmployeeModule());

        System.out.println(mapper.writeValueAsString(emp));
    }
}

参见 Jackson Annotations