如何解析json响应android

问题描述：

如何解析下面的json响应。我正在开发预订应用程序。 i。从服务器得到的响应是如何解析json响应android

{"HotelInformationResponse": { 
"@hotelId": "210985", 
    "customerSessionId": "0ABAA826-9AAF-5791-3692-A03326791310", 
"HotelSummary": { 
    "@order": "0", 
    "hotelId": 210985, 
    "name": "Seattle Marriott Waterfront", 
    "address1": "2100 Alaskan Way", 
    "city": "Seattle", 
    "stateProvinceCode": "WA", 
    "postalCode": 98121, 
    "countryCode": "US", 
    "propertyCategory": 1, 
    "hotelRating": 4, 
    "tripAdvisorRating": 4, 
    "locationDescription": "Near Seattle Aquarium", 
    "highRate": 645, 
    "lowRate": 279, 
    "latitude": 47.61016, 
    "longitude": -122.34651 
}, 
"HotelDetails": { 
    "numberOfRooms": 358, 
    "numberOfFloors": 8, 
    "checkInTime": "4:00 PM", 
    "checkOutTime": "12:00 PM", 
    "propertyInformation": "Pets not allowed Check-in time starts at 4 PM Check-out time is Noon ", 
    }

任何方法被理解

http://stackoverflow.com/questions/2818697/sending-和解析-json-in-android – 2012-04-10 11:54:06

使用JSON，也许？ – m0skit0 2012-04-10 12:02:12

答

你可以使用JSONArray和JSONObject的手动解析它。但它很无聊，在循环运行时可能需要额外注意按键。最简单和强烈推荐的方法是使用Google Gson库自动处理。

答

private JSONObject jObject; 
jObject = new JSONObject(your response as string); 
//this will give you json object for HotelInformationResponse 
JSONObject menuObject = jObject.getJSONObject("HotelInformationResponse"); 
by this way you can get values 
String hotelId = menuObject.getString("@hotelId"); 
String customerSessionId = menuObject.getString("customerSessionId"); 
//...rest are same do same thing for HotelDetails

额外信息Check this link

尽管目标网址上的信息可能会回答问题，但我们更愿意将答案放在我们网站*上，并提供链接作为进一步参考。请在这里包含您的来源的相关信息。 [阅读更多元。]（http://meta.stackexchange.com/a/8259） – animuson 2012-04-11 00:04:19

@animuson - 谢谢你的建议。 – 2012-04-11 04:43:11

答

我希望这个代码可以帮助ü

public static String Profile_response(String response){ 
      try{ 
       JSONArray jsonarray = new JSONArray("["+response+"]"); 
       JSONObject jsonobject = jsonarray.getJSONObject(0); 
       parseForcitydetail1(jsonobject.getString("HotelInformationResponse")); 


       return response; 

      }catch (Exception e) { 
       return "\"" + "Access Denied" + "\""; 
      } 
     } 

    public static void parseForcitydetail1(String res){ 
      try{ 
       JSONArray jsonarray1 = new JSONArray(res); 
       for(int i=0;i<jsonarray1.length();i++){ 
         JSONObject jsonobject = jsonarray1.getJSONObject(i); 
         Hotal_ID.add(jsonobject.getString("@hotelId")); 
         customer_ID.add(jsonobject.getString("customerSessionId")); 

      } 
parseForcitydetail2(jsonobject.getString("HotelSummary")); 


      }catch (Exception e) { 
       System.out.println(e.toString()); 
      } 
     } 

    public static void parseForcitydetail2(String res){ 
      try{ 
       JSONArray jsonarray1 = new JSONArray(res); 
       for(int i=0;i<jsonarray1.length();i++){ 
         JSONObject jsonobject = jsonarray1.getJSONObject(i); 
         Order.add(jsonobject.getString("@order")); 
         hote_ID.add(jsonobject.getString("hotelId")); 
    Name.add(jsonobject.getString("name"));      Address.add(jsonobject.getString("address1"));.... 
    City.add(jsonobject.getString("city")); 
    StateProvinceCode.add(jsonobject.getString("stateProvinceCode")); PostalCode.add(jsonobject.getString("postalCode")); 
    CountryCode.add(jsonobject.getString("countryCode")); 
    PropertyCategory.add(jsonobject.getString("propertyCategory"));      HotelRating.add(jsonobject.getString("hotelRating"));.... 
    TripAdvisorRating.add(jsonobject.getString("tripAdvisorRating")); 
    LocationDescription.add(jsonobject.getString("locationDescription")); Latitude.add(jsonobject.getString("latitude")); 
    Longitude.add(jsonobject.getString("longitude")); 
      } 

      }catch (Exception e) { 
       System.out.println(e.toString()); 
      } 
     }

这个同样的过程 “HotelDetails” 分析

享受！

这是一些可怕的变量命名。 – cdmckay 2012-08-02 13:59:44

答

Using JSONOBject, JSONArray to contain data like JSONObject, JSONArray 
Using methods like: getJSONObject(), getString(), getXXX(). . to get data which you want. 
with XXX - data type like int, string

您将了解如何解析JSON在Android中轻松地描述了如何解析JSON清楚（例）链接： http://www.technotalkative.com/android-json-parsing/

如何解析json响应android

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