运行此代码以获取JSON时，我得到了public void onReceive(Context context, Intent intent) {
NotificationViewModel viewModel =
ViewModelProviders.of(XXX).get(NotificationViewModel.class);
MessageNotification notification =
viewModel.getNextNotification().observe(XXX, new
Observer<MessageNotification>() {
@Override
public void onChanged(MessageNotification messageNotification) {
//do stuff
}
});
}
。但是，当我使用时：
W/System.err: org.json.JSONException: Value（{https://api.myjson.com/bins/j5f6b），一个测试器JSON URL，它为我提供输出。
我尝试在某些地方将其更改为JSON Object，但这无济于事。
[{"name":"Abhishek","password":"123","contact":"1111111111","country":"India"},{"name":"Rahul","password":"1s","contact":"1sdfsdf","country":"India"},{"name":"Abhishek","password":"aar","contact":"asdbsfg","country":"India"}]
我希望看到一个名称输出，除了protected Void doInBackground(Void... voids) {
try {
URL url = new URL("http://www.free-map.org.uk/fm/ws/bsvr.php?
bbox=-0.73,51.04,-0.71,51.06&way=highway&format=json");
HttpURLConnection httpURLConnection = (HttpURLConnection)
url.openConnection();
InputStream inputStream = httpURLConnection.getInputStream();
BufferedReader bufferedReader = new BufferedReader(new
InputStreamReader(inputStream));
String line = "";
while(line != null){
line = bufferedReader.readLine();
data = data + line;
}
JSONArray JA = new JSONArray(data);
for(int i =0 ;i <JA.length(); i++){
JSONObject JO = (JSONObject) JA.get(i);
singleParsed = "name:" + JO.get("name") + "\n";
dataParsed = dataParsed + singleParsed +**"\n"** ;
}
} catch (MalformedURLException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (JSONException e) {
e.printStackTrace();
}
return null;
}
protected void onPostExecute(Void aVoid) {
super.onPostExecute(aVoid);
twod.data.setText(this.dataParsed);
}
答案 0 :(得分：0)
问题似乎出在您的解析逻辑上。
检查：
JSONObject jsonObject = new JSONObject(data);
JSONArray JA = jsonObject.getJSONArray("features");
for (int i = 0; i < JA.length(); i++) {
JSONObject JO = (JSONObject) JA.get(i);
System.out.print(JO);
}
您必须先创建一个JSONObject，然后再从中提取JSONArray。