java - 如何使用 Jackson 实现动态部分 JSON 序列化？

Question

我有一个 POJO，其中包含所有属性的自定义 setter 方法，用于跟踪属性是否已显式设置。 setter 存储到 fieldNameSet boolean 字段，并公开这些标志的 isFieldNameSet getter。 我希望 Jackson 仅使用 isFieldNameSet 为 true 的字段动态序列化该类。

背景:

我开始编写自定义 JsonFilter 实现，但它没有提供有关当前正在序列化的对象实例的任何上下文，因此显然我无法读取 isFieldNameSet 属性的当前值。

Answer 1

从 Jackson 示例中快速破解

import org.codehaus.jackson.JsonGenerationException;
import org.codehaus.jackson.map.JsonMappingException;
import org.codehaus.jackson.map.ObjectMapper;
import org.codehaus.jackson.map.ser.FilterProvider;
import org.codehaus.jackson.map.ser.impl.SimpleBeanPropertyFilter;
import org.codehaus.jackson.map.ser.impl.SimpleFilterProvider;

public class JacksonExample {
    public static void main(String[] args) {

        ObjectMapper mapper = new ObjectMapper();

        User user = createDummyUser();

        try {

            //Its age here , this is conditional based on your fieldset
            SimpleBeanPropertyFilter theFilter = SimpleBeanPropertyFilter.serializeAllExcept("age");
            FilterProvider filters = new SimpleFilterProvider().addFilter("myFilter", theFilter);

            // Convert object to JSON string
            String jsonInString = jsonInString = mapper.writer(filters).writeValueAsString(user);
            System.out.println(jsonInString);

            // Convert object to JSON string and pretty print

            //System.out.println(jsonInString);

        } catch (JsonGenerationException e) {
            e.printStackTrace();
        } catch (JsonMappingException e) {
            e.printStackTrace();
        } catch (IOException e) {
            e.printStackTrace();
        }

    }

    private static User createDummyUser() {

        User user = new User();

        user.setName("mkyong");
        user.setAge(33);

        List<String> msg = new ArrayList<>();
        msg.add("hello jackson 1");
        msg.add("hello jackson 2");
        msg.add("hello jackson 3");

        user.setMessages(msg);

        return user;

    }
}


package org.soproject;

import java.util.List;

import org.codehaus.jackson.map.annotate.JsonFilter;

@JsonFilter("myFilter")
public class User {

    private String name;
    private int age;
    private List<String> messages;

    public String getName() {
        return name;
    }

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

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }

    public List<String> getMessages() {
        return messages;
    }

    public void setMessages(List<String> messages) {
        this.messages = messages;
    }

    // getters and setters

}

忽略年龄，如您所见:

{"name":"mkyong","messages":["你好 jackson 1","你好 jackson 2","你好 jackson 3"]}

注意 jackson 来源来自:https://www.mkyong.com/java/how-to-convert-java-object-to-from-json-jackson/