Json 字符串到 Java Object 带动态键名

Question

我正在尝试将这个结构化的 json 字符串解析为 Java Object 但我每次尝试都失败了。

{
  "message": "Test Message",
  "status": true,
  "users": {
    "user_xy": [
      {
        "time": "2016-08-25 19:01:20.944614158 +0300 EEST",
        "age": 24,
        "props": {
          "pr1": 197,
          "pr2": 0.75,
          "pr3": 0.14,
          "pr4": -0.97
        }
      }
    ],
    "user_zt": [
      {
        "time": "2016-08-25 17:08:36.920891187 +0300 EEST",
        "age": 29,
        "props": {
          "pr1": 1.2332131860505051,
          "pr2": -0.6628148829634317,
          "pr3": -0.11622442112006928
        }
      }
    ]
  }
}

props 字段可以包含 1 个属性或 6 个属性，这取决于数据库记录。 用户名部分也动态变化。

我可以用 Jackson Lib 成功解析这个结构化字符串吗？

Answer 1

您必须创建像波纹管这样的calss结构才能将字符串映射到java对象。

为详细信息创建一个类

public class Details {
    private String message;
    private String status;
    private Map<String, List<UserDetails>> users = new HashMap<String, List<UserDetails>>();
    public String getMessage() {
        return message;
    }
    public void setMessage(String message) {
        this.message = message;
    }
    public String getStatus() {
        return status;
    }
    public void setStatus(String status) {
        this.status = status;
    }
    public Map<String, List<UserDetails>> getUsers() {
        return users;
    }
    public void setUsers(Map<String, List<UserDetails>> users) {
        this.users = users;
    }
}

创建像下面这样的UserDetails类。

public class UserDetails {
    private String time;
    private String age;
    private Map<String, String> prop = new HashMap<String, String>();
    public String getTime() {
        return time;
    }
    public void setTime(String time) {
        this.time = time;
    }
    public String getAge() {
        return age;
    }
    public void setAge(String age) {
        this.age = age;
    }
    public Map<String, String> getProp() {
        return prop;
    }
    public void setProp(Map<String, String> prop) {
        this.prop = prop;
    }   
}

现在，您可以使用Details类映射字符串。

希望这会有所帮助..

Answer 2

 public class Details { private String message; private boolean status; Users UsersObject; // Getter Methods public String getMessage() { return message; } public boolean getStatus() { return status; } public Users getUsers() { return UsersObject; } // Setter Methods public void setMessage(String message) { this.message = message; } public void setStatus(boolean status) { this.status = status; } public void setUsers(Users usersObject) { this.UsersObject = usersObject; } } public class Users { ArrayList < Object > user_xy = new ArrayList < Object > (); ArrayList < Object > user_zt = new ArrayList < Object > (); // Getter Methods // Setter Methods }

也可以在线使用 JSON 转 JAVA Object 转换器： https://codebeautify.org/json-to-java-converter