我正在构建一个Android应用程序,我想要读取我所在城市的当前温度。我正在使用免费的http://openweathermap.org/current API。从JSON获取温度
现在,当我当我尝试这样做,以获取温度以下几点:
@Override
public void onResponse(JSONObject response) {
try {
Log.d("MY LOGGER: ", "Task started");
//Fetch description
JSONArray measurements = response.getJSONArray("list");
String desc = measurements.getJSONObject(0).getJSONArray("weather").getJSONObject(0).getString("description");
//Fetch temp
JSONObject currentTemp = response.getJSONObject("main");
double temp = currentTemp.getDouble("temp");
.
.
的“递减”或描述如预期牵强,买的温度取说:
W/System.err: org.json.JSONException: No value for main W/System.err: at org.json.JSONObject.get(JSONObject.java:389) W/System.err: at org.json.JSONObject.getJSONObject(JSONObject.java:609)
这是我收到的JSON的一个片段,注意 “LON”:XXX “LAT”:xxx是故意的:
{"city":{"id":2624652,"name":"Arhus","coord":{"lon":xxx,"lat":xxx},"country":"DK","population":0,"sys":{"population":0}},"cod":"200","message":0.0028,"cnt":40,"list":[{"dt":1475269200,"main":{"temp":284.9,"temp_min":284.897,"temp_max":284.9,"pressure":1017.03,"sea_level":1020.8,"grnd_level":1017.03,"humidity":79,"temp_kf":0},"weather":[{"id":500,"main":"Rain","description":"light rain","icon":"10n"}],"clouds":{"all":48},"wind":{"speed":6.75,"deg":247.005},"rain":{"3h":0.04},"sys":{"pod":"n"},"dt_txt":"2016-09-30 21:00:00"},{"dt":1475280000,"main":{"temp":284.21,"temp_min":284.205,"temp_max":284.21,"pressure":1017.5,"sea_level":1021.27,"grnd_level":1017.5,"humidity":87,"temp_kf":0},"weather":[{"id":500,"main":"Rain","description":"light rain","icon":"10n"}] ...
我在抓取中做错了什么?
您的JSON是无效的,请检查一下,原来这就是JSON或您张贴目的编辑,你必须定义为XXX –
经/纬度的JSON是我收到的JSON的剪断,说明“LON “:xxx,”lat“:xxx是故意的。 ;-) – Jakob