Java 如何反序列化此JSON
这是我想要解析的JSON数据,但在反序列化方面还是新手,我想从这个JSON数据中获得以下属性: 1.机场代码 2.Latutude 3.经度 这是包含两个机场列表的JSONJava 如何反序列化此JSON,java,android,json,Java,Android,Json,这是我想要解析的JSON数据,但在反序列化方面还是新手,我想从这个JSON数据中获得以下属性: 1.机场代码 2.Latutude 3.经度 这是包含两个机场列表的JSON { "AirportResource": { "Airports": { "Airport": [{ "AirportCode": "AAL", "Position": {
{
"AirportResource": {
"Airports": {
"Airport": [{
"AirportCode": "AAL",
"Position": {
"Coordinate": {
"Latitude": 57.09305556,
"Longitude": 9.85
}
},
"CityCode": "AAL",
"CountryCode": "DK",
"LocationType": "Airport",
"Names": {
"Name": [{
"@LanguageCode": "xx",
"$": "Aalborg"
}, {
"@LanguageCode": "de",
"$": "Aalborg"
}, {
"@LanguageCode": "ru",
"$": "Ольборг"
}, {
"@LanguageCode": "pt",
"$": "Aalborg"
}, {
"@LanguageCode": "jp",
"$": "オールボア"
}, {
"@LanguageCode": "kr",
"$": "올보르그"
}, {
"@LanguageCode": "en",
"$": "Aalborg"
}, {
"@LanguageCode": "it",
"$": "Aalborg"
}, {
"@LanguageCode": "fr",
"$": "Aalborg"
}, {
"@LanguageCode": "es",
"$": "Aalborg"
}, {
"@LanguageCode": "ka",
"$": "奧爾堡"
}, {
"@LanguageCode": "pl",
"$": "Aalbork"
}, {
"@LanguageCode": "mi",
"$": "奥尔堡"
}]
},
"UtcOffset": 2,
"TimeZoneId": "Europe\/Copenhagen"
}, {
"AirportCode": "AAR",
"Position": {
"Coordinate": {
"Latitude": 56.30388889,
"Longitude": 10.62
}
},
"CityCode": "AAR",
"CountryCode": "DK",
"LocationType": "Airport",
"Names": {
"Name": [{
"@LanguageCode": "xx",
"$": "Aarhus"
}, {
"@LanguageCode": "en",
"$": "Aarhus"
}, {
"@LanguageCode": "de",
"$": "Aarhus"
}, {
"@LanguageCode": "it",
"$": "Aarhus"
}, {
"@LanguageCode": "fr",
"$": "Aarhus"
}, {
"@LanguageCode": "es",
"$": "Aarhus"
}]
},
"UtcOffset": 2,
"TimeZoneId": "Europe\/Copenhagen"
}]
},
"Meta": {
"@Version": "1.0.0",
"Link": [{
"@Href": "https:\/\/api.lufthansa.com\/v1\/references\/airports\/?limit=2&LHoperated=0&offset=0",
"@Rel": "self"
}, {
"@Href": "https:\/\/api.lufthansa.com\/v1\/references\/airports\/?limit=2&LHoperated=0&offset=2",
"@Rel": "next"
}, {
"@Href": "https:\/\/api.lufthansa.com\/v1\/references\/airports\/?limit=2&LHoperated=0&offset=1260",
"@Rel": "last"
}, {
"@Href": "https:\/\/api.lufthansa.com\/v1\/references\/cities\/{cityCode}",
"@Rel": "related"
}, {
"@Href": "https:\/\/api.lufthansa.com\/v1\/references\/countries\/{countryCode}",
"@Rel": "related"
}],
"TotalCount": 1261
}
}
}
然后我的Javacode:
编辑:
private void parseJson(String result)
{
try {
if (result!=null) {
JSONObject obj = new JSONObject(result).getJSONObject("AirportResource").getJSONObject("Airports");
JSONArray arr = obj.getJSONArray("Airport");
for (int i = 0; i < arr.length(); i++)
{
String AirportCode = arr.getJSONObject(i).getString("AAL");
String Latitude = arr.getJSONObject(i).getJSONObject("Position").getJSONObject("Coordinate").getString("Latitude");
String Longitude = arr.getJSONObject(i).getJSONObject("Position").getJSONObject("Coordinate").getString("Longitude");
System.out.println("Airport : " + AirportCode + " Latitude: " + Latitude+
" Longitude : " + Longitude );
}
}
else{
Toast.makeText(MainActivity.this,Config.POOR_NETWORK_CONNECTION, Toast.LENGTH_LONG).show();
}
}
catch (JSONException r){
System.out.println("ERROR PROB : "+ r);
}
}
如何解析这种JSON类型的数据?您没有使用Airports对象。在Airports内部对象是Airport数组,您可以直接访问它。因此抛出错误
JSONObject obj2 = obj.getJSONObject("Airports");
现在从obj获取Json数组
JsonArray jsonarray= obj2.getJsonArray("Airport");
现在就在jsonarray的长度上使用for循环。我建议使用它来正确查看和格式化JSON
这就是结果
从这里你可以很容易地解决你的问题
JSONObject AirportResource= obj.getJSONObject("AirportResource");
JsonArray AirPorts= AirportResource.getJSONObject("Airports");
JsonArray jsonarray= AirPorts.getJsonArray("Airport");
for (int i = 0; i < jsonarray.length(); i++) {
JSONObject obj= AirportResource.get(i); do sth with it
obj.getString("key"); //do sth with it
}
JSONObject AirportResource=obj.getJSONObject(“AirportResource”);
JsonArray AirPorts=AirportResource.getJSONObject(“机场”);
JsonArray JsonArray=AirPorts.getJsonArray(“Airport”);
for(int i=0;i
这将有望让您开始学习
private static void parseJson(JSONObject object) throws JSONException {
if (object!=null) {
JSONObject airportResource = object.getJSONObject("AirportResource");
JSONObject airports = airportResource.getJSONObject("Airports");
JSONArray airportArray = airports.getJSONArray("Airport");
for (int i = 0 ; i < airportArray.length(); i++) {
JSONObject airport = (JSONObject) airportArray.get(i);
//System.out.println(airport);
System.out.println("CityCode: " + airport.getString("CityCode"));
}
}
else{
//null
}
}
private static void parseJson(JSONObject对象)抛出JSONException{
if(对象!=null){
JSONObject airportResource=object.getJSONObject(“airportResource”);
JSONObject airports=airportResource.getJSONObject(“机场”);
JSONArray airportArray=airports.getJSONArray(“机场”);
对于(int i=0;i
输出将是:
城市代码:AAL
城市代码:AAR
您似乎忘记了代码中的
Airports
对象AirportResource(object)->Airports(object)->Airport(array)
@xander,请帮助我编辑我建议您为上面的JSON创建POJO,并使用诸如Gson之类的库来解析它。您可以排除JSON字符串中的无关数据。然后解析将像gson.fromJson(json,AirportResource.class)一样简单代码>。如果记录结果
您会得到什么?请尝试System.out.println(result)
您得到了什么?这更多的是一个评论而不是一个答案。您好,请尝试将我的代码编辑为您的答案编辑它。我想,Op应该自己尝试编码。您好,我已经将我的代码编辑为JSONObject obj=new JSONObject(result)。getJSONObject(“AirportResource”)。getJSONObject(“Airports”)代码>不要只是复制和粘贴代码。想想这个解。如果你不知道代码要做什么,你就永远不会正确编码。但是代码中有一些错误。我会给你一个答案。这和他在问题中编辑的代码有什么不同?另外,JsonArray机场
的类型应为JSONObject
,但我想这是一个复制错误。:)我以为他对JSON结构不太了解,我又看了一遍问题并更新了答案。谢谢您的反馈。@xander:)
private static void parseJson(JSONObject object) throws JSONException {
if (object!=null) {
JSONObject airportResource = object.getJSONObject("AirportResource");
JSONObject airports = airportResource.getJSONObject("Airports");
JSONArray airportArray = airports.getJSONArray("Airport");
for (int i = 0 ; i < airportArray.length(); i++) {
JSONObject airport = (JSONObject) airportArray.get(i);
//System.out.println(airport);
System.out.println("CityCode: " + airport.getString("CityCode"));
}
}
else{
//null
}
}