java - @JsonView 在简单测试中不起作用

Question

我无法使用@JsonView 执行这个简单的示例。我做错了什么？

${jackson-2-version} = 2.6.5

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-core</artifactId>
    <version>${jackson-2-version}</version>
</dependency>

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-annotations</artifactId>
    <version>${jackson-2-version}</version>
</dependency>

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>${jackson-2-version}</version>
</dependency>

完整的测试类。

package staticTest;

import com.fasterxml.jackson.annotation.JsonView;
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.databind.ObjectMapper;

import org.junit.Test;

/**
 * Created by Daniel on 01/04/2016.
 */
public class Jackson2Tests {

    @Test
    public void JsonViewTest(){

        try {

            System.out.println(getMapper().writeValueAsString(new DemoClass()));
        } catch (JsonProcessingException e) {

            e.printStackTrace();
        }
    }

    private ObjectMapper getMapper(){

        ObjectMapper objectMapper = new ObjectMapper();

        objectMapper.writerWithView(ToShowIn.App.class);
        objectMapper.readerWithView(ToShowIn.App.class);

        return objectMapper;
    }

    public class ToShowIn {

        public class App{}
        public class Manager{}

    }

    class DemoClass{

        @JsonView(ToShowIn.App.class)
        private String propertyOne = "one";
        private int propertyTwo = 2;
        private boolean propertyThree = true;
        private DemoChild propertyFour = new DemoChild();

        public String getPropertyOne() {
            return propertyOne;
        }

        public void setPropertyOne(String propertyOne) {
            this.propertyOne = propertyOne;
        }

        public int getPropertyTwo() {
            return propertyTwo;
        }

        public void setPropertyTwo(int propertyTwo) {
            this.propertyTwo = propertyTwo;
        }

        public boolean isPropertyThree() {
            return propertyThree;
        }

        public void setPropertyThree(boolean propertyThree) {
            this.propertyThree = propertyThree;
        }

        public DemoChild getPropertyFour() {
            return propertyFour;
        }

        public void setPropertyFour(DemoChild propertyFour) {
            this.propertyFour = propertyFour;
        }

        class DemoChild{

            private String childOne = "1";
            private int childTwo = 2;
            private boolean childThree = true;

            public String getChildOne() {
                return childOne;
            }

            public void setChildOne(String childOne) {
                this.childOne = childOne;
            }

            public int getChildTwo() {
                return childTwo;
            }

            public void setChildTwo(int childTwo) {
                this.childTwo = childTwo;
            }

            public boolean isChildThree() {
                return childThree;
            }

            public void setChildThree(boolean childThree) {
                this.childThree = childThree;
            }
        }

    }

}

结果:

{"propertyOne":"one","propertyTwo":2,"propertyThree":true,"propertyFour":{"childOne":"1","childTwo":2,"childThree":true}}

解决方案

首先，正如@varren 在下面所说，.writerWithView() 不是 setter。

要在 ObjectMapper 中使用 .setConfig() 使一个 View 持久化。

objectMapper.setConfig(objectMapper.getSerializationConfig().withView(ToShowIn.App.class));

最后，也是最重要的，objectMapper.disable(MapperFeature.DEFAULT_VIEW_INCLUSION);让 JsonView 的工作方式就像@varren 贡献的 jackson 文档所说的那样。

Answer 1

writerWithView 和 readerWithView 不是 setter，它们是 ObjectWriter 和 ObjectReader 构建器方法，所以你必须使用:

getMapper().writerWithView(ToShowIn.App.class).writeValueAsString(new DemoClass()))

而且您还必须为您的 objectMapper 禁用 MapperFeature.DEFAULT_VIEW_INCLUSION。

private ObjectMapper getMapper(){
    ObjectMapper objectMapper = new ObjectMapper();
    objectMapper.disable(MapperFeature.DEFAULT_VIEW_INCLUSION);
    return objectMapper;
}

这是文档中的信息:http://wiki.fasterxml.com/JacksonJsonViews

Handling of "view-less" properties

By default all properties without explicit view definition are included in serialization. But starting with Jackson 1.5 you can change this default by:

objectMapper.configure(SerializationConfig.Feature.DEFAULT_VIEW_INCLUSION, false); where false means that such properties are NOT included when enabling with a view. Default for this property is 'true'.