Question

这些是我的课程。

Class TypeC {

    int var1;
    HashMap<String,String>var2;
    ArrayList<TypeC> var3;

}

Class TypeB {

TypeC var1;

}

Class TypeA {

    Long var1;
    TypeB var2;
}

我想创建TypeC的对象，然后将其转换为相应的JSON对象（复杂的JSON）。我尝试了以下但它不起作用。

    TypeC obj = new TypeC();
    JSONObject TypeCJSON=new JSONObject(obj);

Answer 1

使用'com.fasterxml.jackson.databind.ObjectMapper'进行数据绑定的完整示例：

package spring.exos;

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

public class Main {

public static void main(String[] args){

    final Computer computer = new Computer();
    computer.setBrand("Toshiba");
    computer.setModel("TSB I7-SSD");
    computer.setSpecs(new Specs(new Integer(256), new Integer(8), new Double(2.4)));

    final ObjectMapper mapper = new ObjectMapper();
    try {
        System.out.println(mapper.writeValueAsString(computer));
    } catch (JsonProcessingException e) {
        e.printStackTrace();
    }

}

public static class Computer{
    private String brand;
    private String model;
    private Specs specs;

    public String getBrand() {
        return brand;
    }
    public void setBrand(String brand) {
        this.brand = brand;
    }
    public String getModel() {
        return model;
    }
    public void setModel(String model) {
        this.model = model;
    }
    public Specs getSpecs() {
        return specs;
    }
    public void setSpecs(Specs specs) {
        this.specs = specs;
    }
}
public static class Specs {
    private Integer hdd;
    private Integer memory;
    private Double cpu;

    public Specs(Integer hdd, Integer memory, Double cpu) {
        super();
        this.hdd = hdd;
        this.memory = memory;
        this.cpu = cpu;
    }
    public Integer getHdd() {
        return hdd;
    }
    public void setHdd(Integer hdd) {
        this.hdd = hdd;
    }
    public Integer getMemory() {
        return this.memory;
    }
    public void setMemory(Integer memory) {
        this.memory = memory;
    }
    public Double getCpu() {
        return cpu;
    }
    public void setCpu(Double cpu) {
        this.cpu = cpu;
    }
}
}

输出结果为：

{“brand”：“Toshiba”，“model”：“TSB I7-SSD”，“specs”：{“hdd”：256，“memory”：8，“cpu”：2.4}}

您需要依赖：

<dependency>
    <groupId>com.fasterxml.jackson.core</groupId>
    <artifactId>jackson-databind</artifactId>
    <version>2.7.1-1</version>
</dependency>

Answer 2

如果您愿意使用其他库，请使用Gson https://github.com/google/gson，您只需要这样做：

<itemizedlist role="type6" id="ecls_bio_becls_a3_a43205230.SL2392155.512">

从java对象创建复杂的JSON

2 个答案: