Question

如何转换此json

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

使用Jackson mixin

获取员工对象

//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，您会在序列化时将address字段注释为@JsonUnwrapped并将其解包（并在反序列化时包装）。但是，由于您无法更改第三方课程，因此您必须在mixin上执行此操作：

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

然后，创建一个包含所有特定＆＃34;扩展名＆＃34;的模块。这可以通过继承Module或SimpleModule或通过撰写SimpleModule来完成：

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

第三，使用ObjectMapper注册模块：

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

最后，有趣的序列化/反序列化！

子类SimpleModule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));
    }
}

请参阅Jackson Annotations

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

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