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java - 如何将包含自定义对象列表的地图字段的自定义对象转换为 json？

问题描述

假设，我有一个 Employee 类，如下所示：

public class Employee{

Map<String, ArrayList<Salary>> salary = new HashMap<String, ArrayList<Salary>>();
String name;
String age;
}

public class Salary{
String amount;
String currency;
}

在 Java 中与 Json 进行转换的最聪明的方法是什么？

或者;

如果我的 json 看起来像这样怎么办：

  {
  "name": "Test",
  "age": "12",
  "salary": {
    "first": {
      "41130": {
        "amount": "100",
        "currency": "€"
      },
      "41132": {
        "amount": "100",
        "currency": "€"
      }
    },
    "second": {
      "41129": {
        "amount": "100",
        "currency": "€"
      }
    }
  }
}

当我尝试将其转换为 Employee 时，出现以下错误。

com.google.gson.JsonSyntaxException：java.lang.IllegalStateException：应为 BEGIN_ARRAY 但为 BEGIN_OBJECT

标签： javajsonjacksongson

解决方案

public class Main {

    public static void main(String[] args) {

        Gson gson = new Gson();

        Map<String, ArrayList<Salary>> sal = new HashMap<String, ArrayList<Salary>>();
        ArrayList<Salary> salaries = new ArrayList<Salary>();
        Salary salary1 = new Salary("100", "€");
        Salary salary2 = new Salary("200", "€");
        salaries.add(salary1);
        salaries.add(salary2);
        sal.put("1", salaries);
        Employee employee = new Employee(sal, "Test", "12");

        System.out.println("Age -> " + employee.getAge());
        System.out.println("Name -> " + employee.getName());
        System.out.println("Salary -> " + employee.getSalary());

        String json = gson.toJson(employee);
        System.out.println("Json -> " + json);

        Employee employee1 = gson.fromJson(json, Employee.class);

        System.out.println("Age1 -> " + employee1.getAge());
        System.out.println("Name1 -> " + employee1.getName());
        System.out.println("Salary1 -> " + employee1.getSalary());
    }

    @Data
    @AllArgsConstructor
    @NoArgsConstructor
    public static class Employee{

        Map<String, ArrayList<Salary>> salary = new HashMap<String, ArrayList<Salary>>();
        String name;
        String age;
    }

    @Data
    @AllArgsConstructor
    @NoArgsConstructor
    public static class Salary{
        String amount;
        String currency;

    }
}

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