使用Jackson和Mockito序列化对象时的无限递归

Question

我在使用MockMvc,Mockito和Jackson测试Spring控制器时遇到了这个问题,所以我做了一个简单的类来测试杰克逊的表现.我使用的是jackson-databind:2.3.1和mockito-core:1.9.5.

鉴于此类:

import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.ObjectMapper;
import java.io.Serializable;
import static org.mockito.Mockito.mock;
import static org.mockito.Mockito.when;

public class Person implements Serializable {
    private String name;
    private int age;

    // Public getters and setters...

    public static void main(String[] args) {
        String name = "Bob";
        int age = 21;
        ObjectMapper objectMapper = new ObjectMapper();

        // attempt serialization with real object
        Person person = new Person();
        person.setName(name);
        person.setAge(age);
        try {
            System.out.println(objectMapper.writeValueAsString(person));
        } catch (JsonProcessingException e) {
            e.printStackTrace();
            System.err.println("Failed to serialize real object");
        }

        // attempt serialization with mock object
        Person mockPerson = mock(Person.class);
        when(mockPerson.getName()).thenReturn(name);
        when(mockPerson.getAge()).thenReturn(age);
        try {
            System.out.println(objectMapper.writeValueAsString(mockPerson));
        } catch (JsonProcessingException e) {
            e.printStackTrace();
            System.err.println("Failed to serialize mock object.");
        }
    }

Jackson在序列化真实对象方面没有问题,但是当它尝试序列化模拟对象时会抛出JsonMappingException.通过代码调试,它反复调用serializeFields(bean,jgen,provider),卡在内部Mockito属性上.

所以,我的问题是:无论如何迫使杰克逊使用getter方法？我在类上尝试了@JsonIgnoreProperties,在字段上尝试了@JsonIgnore,在方法上尝试了@JsonProperty(在不同的组合中,没有成功).或者,我是否必须编写自己的自定义序列化程序？

谢谢!

Answer 1

这是一个适合您特定情况的解决方案:

首先,您需要创建一个PersonMixin,因为您无法将所需的注释添加到模拟中.

import com.fasterxml.jackson.annotation.JsonAutoDetect;
import com.fasterxml.jackson.annotation.JsonProperty;


@JsonAutoDetect(getterVisibility = JsonAutoDetect.Visibility.NONE, isGetterVisibility = JsonAutoDetect.Visibility.NONE)
public interface PersonMixin {

    @JsonProperty
    String getName();

    @JsonProperty
    Integer getAge();
}

现在,使用对象映射器,如下面的代码,您将获得与序列化真实对象时相同的结果:

Person mockPerson = mock(Person.class);
when(mockPerson.getName()).thenReturn(name);
when(mockPerson.getAge()).thenReturn(age);
objectMapper.addMixInAnnotations(Person.class, PersonMixin.class);
try {
    System.out.println(objectMapper.writeValueAsString(mockPerson));
} catch (JsonProcessingException e) {
    e.printStackTrace();
    System.err.println("Failed to serialize mock object.");
}

Answer 2

这是我ObjectMapper在不需要 mixin 的情况下将其整理出来的。

映射器会忽略名称中某处带有“Mockito”的所有成员。

此解决方案避免了每个序列化对象的混合，或注释可能无法访问的代码。

运行以下测试成功并输出{"name":"Jonh"}。

package test;

import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.ObjectMapper;
import com.fasterxml.jackson.databind.introspect.AnnotatedMember;
import com.fasterxml.jackson.databind.introspect.JacksonAnnotationIntrospector;
import org.mockito.Mockito;

public class AppTest extends Mockito {

    public void testApp() throws JsonProcessingException {
        ObjectMapper mapper = new ObjectMapper();
        mapper.setAnnotationIntrospector(new JacksonAnnotationIntrospector() {

            @Override
            public boolean hasIgnoreMarker(final AnnotatedMember m) {
                return super.hasIgnoreMarker(m) || m.getName().contains("Mockito");
            }
        });

        final String name = "Jonh";

        Person mockPerson = mock(Person.class);
        when(mockPerson.getName()).thenReturn(name);
        System.out.println(mapper.writeValueAsString(mockPerson));
    }


    public static class Person {

        private String name;

        public String getName() {
            return name;
        }

        public void setName(String name) {
            this.name = name;
        }
    }
}