带有动态密钥名称的Json String to Java Object [英] Json String to Java Object with dynamic key name

问题描述

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

I'm trying to parse this structured of json string to Java Object but I failed every attempt.

{
  "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
        }
      }
    ]
  }
}

道具字段可以包含1个属性或6个属性，它取决于db记录。用户名部分也动态变化。

props field can contain 1 properties or 6 properties, it depends on db record. Also Username part dynamically changing.

我可以用Jackson Lib成功解析这个结构化字符串吗？

Can I parse successfully this structured string with Jackson Lib?

推荐答案

你必须像bellow一样创建calss结构，将你的字符串映射到java对象。

You have to create calss structure like bellow to map your string to java object.

为Details创建一个类

Create one class for Details

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;
    }
}

像bellow一样创建UserDetails类。

Create UserDetails class like bellow.

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。< Br>

Now you can map your string with Details class.

希望这会有所帮助..

Hope this will help..

这篇关于带有动态密钥名称的Json String to Java Object的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！