用java解析json

Question

我是json解析的新手,我从请求中抓取了一个json字符串,现在我需要用java解析它.我正在使用json-lib.但我真的被困住了,因为我不熟悉它.我需要提取以下数据

1. name (hotel name)
2. starRating
3. geoPoint

我使用了以下java代码,但它没有给我我需要的结果,请有人帮助我...

非常感谢!

java代码(s是我得到的json字符串)

JSONObject json = (JSONObject) JSONSerializer.toJSON(s);    
JSONArray jarray = json.getJSONArray("hotels");
for(int i=0 ; i < jarray.size(); i++) {
System.out.println("jarray [" + i + "] --------" + jarray.getString(i));
}

json我需要解析

[
{
    "total": 250,
    "offset": 0,
    "requestID": "-btygi09oxfov",
    "locationName": "Paris, France",
    "locationLatitude": 48.86,
    "locationLongitude": 2.34,
    "cityCode": "PARIS_J_FR",
    "hotels": [
        {
            "ypid": "YN10001x300073304",
            "id": 56263,
            "hotelRateIndicator": "2",
            "name": "Renaissance Paris Vendome Hotel",
            "brandCode": "69",
            "addressLine1": "4 Rue du Mont-Thabor",
            "city": "Paris",
            "neighborhood": "",
            "state": "IdF",
            "country": "US",
            "cachedPrice": 935,
            "geoPoint": [
                48.865361,
                2.329584
            ],
            "starRating": "5",
            "thumbnailUrl": "http://www.orbitz.com//public/hotelthumbnails/53/97/85397/85397_TBNL_1246535840051.jpg",
            "total": 250,
            "amenities": [
                "24",
                "31",
                "42",
                "52",
                "9"
            ],
            "telephoneNumbers": [
                ""
            ],
            "popularity": 837
        },
        {
            "ypid": "YN10001x300073331",
            "id": 112341,
            "hotelRateIndicator": "3",
            "name": "Renaissance Paris Arc de Triomphe Hotel",
            "brandCode": "69",
            "addressLine1": "39 Avenue de Wagram",
            "city": "Paris",
            "neighborhood": "",
            "state": "IdF",
            "country": "US",
            "cachedPrice": 633,
            "geoPoint": [
                48.877107,
                2.297451
            ],
            "starRating": "5",
            "thumbnailUrl": "http://www.orbitz.com//public/hotelthumbnails/21/72/302172/302172_TBNL_1246535872514.jpg",
            "total": 250,
            "amenities": [
                "24",
                "31",
                "42",
                "9"
            ],
            "telephoneNumbers": [
                ""
            ],
            "popularity": 796
        }           
  ]         
}           
  ]

Answer 1

任何JSON对象都可以表示为Map<String, Object>.

使用像jackson这样的库(随spring一起提供),它可以将json反序列化为Map,如下所示:

Map<String, Object> obj = new ObjectMapper().readValue(json, new TypeReference<Map<String, Object>>());

或者速度较慢但谷歌品牌的GSON可以像.

Map<String, Object> obj = new Gson().fromJson(json, HashMap.class);

Answer 2

要超越ClassCastException,您只需要进行更改,告诉您:将输入作为数组处理而不是作为对象处理.

JSONArray outerArray = (JSONArray) JSONSerializer.toJSON(s);
JSONObject json = (JSONObject) outerArray.get(0);
JSONArray jarray = json.getJSONArray("hotels");
for (int i = 0; i < jarray.size(); i++)
{
  System.out.println("jarray [" + i + "] --------" + jarray.getString(i));
}

并且,这是获取每个酒店名称的示例.

JSONArray outerArray = (JSONArray) JSONSerializer.toJSON(s);
JSONObject json = (JSONObject) outerArray.get(0);
JSONArray jarray = json.getJSONArray("hotels");
for (int i = 0; i < jarray.size(); i++)
{
  JSONObject hotel = jarray.getJSONObject(i);
  String name = hotel.getString("name");
  System.out.println(name);
}