Java 如何迭代JSONObject以获取单个项

Java 如何迭代JSONObject以获取单个项,java,json,Java,Json,这是我下面的代码,我需要从中解析JSONObject,以获取单个项目。这是我第一次使用JSON。因此,不确定如何解析JSONObject以从JSONObject获取单个项 try { String url = service + version + method + ipAddress + format; StringBuilder builder = new StringBuilder(); httpclient = new DefaultHttpClient();

这是我下面的代码,我需要从中解析
JSONObject
,以获取单个项目。这是我第一次使用JSON。因此,不确定如何解析
JSONObject
以从
JSONObject
获取单个项

try {
    String url = service + version + method + ipAddress + format;
    StringBuilder builder = new StringBuilder();
    httpclient = new DefaultHttpClient();
    httpget = new HttpGet(url);
    httpget.getRequestLine();
    response = httpclient.execute(httpget);
    HttpEntity entity = response.getEntity();
    if (entity != null) {
        InputStream inputStream = entity.getContent();
        bufferedReader = new BufferedReader(new InputStreamReader(inputStream));
        for (String line = null; (line = bufferedReader.readLine()) != null;) {
            builder.append(line).append("\n");
        }
        JSONObject jsonObject = new JSONObject(builder.toString());
        // Now iterate jsonObject to get Latitude,Longitude,City,Country etc etc.

    }

} catch (Exception e) {
    getLogger().log(LogLevel.ERROR, e.getMessage());
} finally {
    bufferedReader.close();
    httpclient.getConnectionManager().shutdown();
}
我的JSON如下所示:

{
    "ipinfo": {
        "ip_address": "131.208.128.15",
        "ip_type": "Mapped",
        "Location": {
            "continent": "north america",
            "latitude": 30.1,
            "longitude": -81.714,
            "CountryData": {
                "country": "united states",
                "country_code": "us"
            },
            "region": "southeast",
            "StateData": {
                "state": "florida",
                "state_code": "fl"
            },
            "CityData": {
                "city": "fleming island",
                "postal_code": "32003",
                "time_zone": -5
            }
        }
    }
}
我需要从上述对象获取
纬度
经度
城市
国家
邮政编码
。有人能提供一些有效的建议吗?

这个怎么样

JSONObject jsonObject = new JSONObject           (YOUR_JSON_STRING);
JSONObject ipinfo     = jsonObject.getJSONObject ("ipinfo");
String     ip_address = ipinfo.getString         ("ip_address");
JSONObject location   = ipinfo.getJSONObject     ("Location");
String     latitude   = location.getString       ("latitude");
System.out.println (latitude);

此示例代码使用

您可以尝试使用它,它将递归地查找json对象中的所有键值,并将其构造为映射。您只需从地图上获取所需的密钥即可

public static Map<String,String> parse(JSONObject json , Map<String,String> out) throws JSONException{
    Iterator<String> keys = json.keys();
    while(keys.hasNext()){
        String key = keys.next();
        String val = null;
        try{
             JSONObject value = json.getJSONObject(key);
             parse(value,out);
        }catch(Exception e){
            val = json.getString(key);
        }

        if(val != null){
            out.put(key,val);
        }
    }
    return out;
}

 public static void main(String[] args) throws JSONException {

    String json = "{'ipinfo': {'ip_address': '131.208.128.15','ip_type': 'Mapped','Location': {'continent': 'north america','latitude': 30.1,'longitude': -81.714,'CountryData': {'country': 'united states','country_code': 'us'},'region': 'southeast','StateData': {'state': 'florida','state_code': 'fl'},'CityData': {'city': 'fleming island','postal_code': '32003','time_zone': -5}}}}";

    JSONObject object = new JSONObject(json);

    JSONObject info = object.getJSONObject("ipinfo");

    Map<String,String> out = new HashMap<String, String>();

    parse(info,out);

    String latitude = out.get("latitude");
    String longitude = out.get("longitude");
    String city = out.get("city");
    String state = out.get("state");
    String country = out.get("country");
    String postal = out.get("postal_code");

    System.out.println("Latitude : " + latitude + " LongiTude : " + longitude + " City : "+city + " State : "+ state + " Country : "+country+" postal "+postal);

    System.out.println("ALL VALUE " + out);

}

回答得好。我要提到的是,在这种情况下,
Map
工作得很好,但为了使它更通用,我会使用
Map
,因为有些字段是数字字段。
    Latitude : 30.1 LongiTude : -81.714 City : fleming island State : florida Country : united states postal 32003
ALL VALUE {region=southeast, ip_type=Mapped, state_code=fl, state=florida, country_code=us, city=fleming island, country=united states, time_zone=-5, ip_address=131.208.128.15, postal_code=32003, continent=north america, longitude=-81.714, latitude=30.1}