Java& Json：序列化时包装元素 [英] Java & Json: Wrapping elements when serialising

问题描述

只是想知道是否有办法做到这一点 - 我有一个类，比如

Just wondering if there was a way to do this - I have a class, something like

class MyClass {
    private String name;
    private String address;
    private String number;
}

当我使用Jackson将其序列化为Json时，我想要包装字符串变量在一起，所以它看起来像

When I serialise it to Json, using Jackson, I want to wrap the String variables together, so it would look something like

{
    "Strings": {
         "name" : "value",
         "address" : "value"
    }
 }

没有将这些变量包装在MyClass中的List或Map类中......这可能吗？

Without wrapping those variables in a List or Map class inside the MyClass... is this possible?

推荐答案

你也可以在你的POJO课程中为Strings，Intigers等添加额外的getter。这些方法应该返回 Map 的结果。请考虑以下代码：

You can also add into your POJO class additional getters for "Strings", "Intigers", etc. Those methods should return Map's as a result. Consider below code:

import java.util.LinkedHashMap;
import java.util.Map;

import com.fasterxml.jackson.annotation.JsonIgnore;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.databind.ObjectMapper;

public class JacksonProgram {

    public static void main(String[] args) throws Exception {
        ObjectMapper mapper = new ObjectMapper();
        MyClass myClass = new MyClass();
        myClass.setAddress("New York, Golden St. 1");
        myClass.setName("James Java");
        myClass.setNumber("444");

        System.out.println(mapper.writerWithDefaultPrettyPrinter().writeValueAsString(myClass));
    }
}

class MyClass {

    private String name;
    private String address;
    private String number;

    @JsonIgnore
    public String getName() {
        return name;
    }

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

    @JsonIgnore
    public String getAddress() {
        return address;
    }

    public void setAddress(String address) {
        this.address = address;
    }

    @JsonIgnore
    public String getNumber() {
        return number;
    }

    public void setNumber(String number) {
        this.number = number;
    }

    @JsonProperty(value = "Strings")
    public Map<String, String> getStrings() {
        Map<String, String> map = new LinkedHashMap<String, String>();
        map.put("name", getName());
        map.put("address", getAddress());
        map.put("number", getNumber());
        return map;
    }
}

结果：

{
  "Strings" : {
    "name" : "James Java",
    "address" : "New York, Golden St. 1",
    "number" : "444"
  }
}

这可能是您可以使用的最优雅的解决方案，但它很简单。

This is not, probably, the most elegant solution which you can use, but it is simple.

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