試圖手動JSON轉換爲類對象過程中自動

Question

樣品JSON數據

{ 
    "Status": true, 
    "Message": "Roles retrieved successfully", 
    "Data": [ 
    { 
     "Role": "SuperAdmin", 
    }, 
    { 
     "Role": "Admin", 
    } 
    ] 
} 

角色類

public class Role { 
    public int Role; 

    public int getRole() { 
     return Role; 
    } 

    public void setRoleID(String role) { 
     Role = role; 
    } 

} 

JSON數據等價類

public class RoleData { 
    public String Message; 
    public boolean Status; 
    public List<Role> Data; 


    public String getMessage() { 
     return Message; 
    } 

    public void setMessage(String message) { 
     Message = message; 
    } 

    public boolean isStatus() { 
     return Status; 
    } 

    public void setStatus(boolean status) { 
     Status = status; 
    } 

    public List<Role> getData() { 
     return Data; 
    } 

    public void setData(List<Role> data) { 
     Data = data; 
    } 
} 

下面是我的代碼來解析檢索的JSON數據導入型RoleData的類對象

private void parseJsonResponse(String responsestring) throws JSONException { 

    RoleData roleData = new RoleData(); 
    JSONObject response = new JSONObject(responsestring); 
    if ((response == null) || response.length() == 0) { 
     return; 
    } 
    try { 

     if (response.getBoolean("Status")) { 
      JSONArray data = response.getJSONArray("Data"); 
      List<Role> temp = new ArrayList<Role>(); 
      for (int i = 0; i < data.length(); i++) { 
       JSONObject dataObj = data.getJSONObject(i); 
       Role role = new Role(); 
       role.setRole(dataObj.getString("Role")); 
       roleData.setMessage(response.getString("Message")); 
       roleData.setStatus(response.getBoolean("Status")); 
       temp.add(role); 
      } 
      roleData.setData(temp); 
     } 
    }catch(JSONException e){ 
     e.printStackTrace(); 
    } 
} 

問題

有沒有更好的辦法來避免上面寫着解析代碼？在.NET中，我們可以使用JsonConvert.DeserializeObject方法將JSON轉換爲類對象，如下所示。

var result = JsonConvert.DeserializeObject<RoleData>(JSON_Response); 

Answer 1

結帳這個庫：

https://github.com/google/gson

https://github.com/FasterXML/jackson（支持JSON非常好）

https://www.mkyong.com/java/jackson-2-convert-java-object-to-from-json/

兩者都基於註解配置不相匹配的變量在JSON中的名稱，通常他們不是必需的:)