Question

我尝试过多种方法来解析这些数据：

JSONObject hospital2 = hospital.getJSONObject("address");
if (!hospital2.isNull("hospital")){
    String hospitalName = hospital2.getString("hospital");
    String hospitalLat = hospital.getString("lat");
    String hospitalLon = hospital.getString("lon");

看起来像这样 - ＆gt;＆gt;＆gt; http://imgur.com/a/OloES 从JSON网址，但要么工作

我的代码atm就是这样：

public class HospActivity extends AppCompatActivity {


    public void goBack(View view)
    {
        Intent i = new Intent(getApplicationContext(),MainActivity.class);
        startActivity(i);
    }


    public class DownloadTask extends AsyncTask<String,Void,String> {

        @Override
        protected String doInBackground(String... params) {

            String result="";
            URL url;
            HttpURLConnection urlConnection = null;

            try {
                url = new URL(params[0]);
                urlConnection = (HttpURLConnection) url.openConnection();

                InputStream in = urlConnection.getInputStream();
                InputStreamReader reader = new InputStreamReader(in);

                int data = reader.read();
                while(data != -1){
                    char current = (char) data;
                    result += current;
                    data = reader.read();
                }

                return result;

            } catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }

            return null;
        }
    }

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_hosp);

        ListView listView = (ListView) findViewById(R.id.listView);
        ArrayList<String> hospitals = new ArrayList<String>();



        DownloadTask task = new DownloadTask();
        String result = null;
        try {
            result = task.execute("http://nominatim.openstreetmap.org/search?q=crete%5bhospitals%5d+greece&format=json&polygon=0&addressdetails=1&limit=500").get();
        } catch (InterruptedException e) {
            e.printStackTrace();
        } catch (ExecutionException e) {
            e.printStackTrace();
        }
        try {
            JSONArray array = new JSONArray(result);
            String asef = "";


            for(int i = 0; i < array.length(); i++)
            {
                JSONObject hospital = array.getJSONObject(i);
                JSONObject hospital2 = hospital.getJSONObject("address");
                if (!hospital2.isNull("hospital")){
                    String hospitalName = hospital2.getString("hospital");
                    String hospitalLat = hospital.getString("lat");
                    String hospitalLon = hospital.getString("lon");




                }
            }


        } catch (JSONException e) {
            e.printStackTrace();
        }



    }
}

我是Android开发的新手。

我的JSON，

[{
    "place_id": "146977563",
    "licence": "Data © OpenStreetMap contributors, ODbL 1.0. http:\/\/www.openstreetmap.org\/copyright",
    "osm_type": "way",
    "osm_id": "377010610",
    "boundingbox": ["35.3648199", "35.3654346", "24.4711123", "24.472325"],
    "lat": "35.3650817",
    "lon": "24.4716895296637",
    "display_name": "Koumoundourou, Rethymno, Rethymni Municipality, Rethymno Regional Unit, Region of Crete, Decentralized Administration of Crete, Greece",
    "class": "amenity",
    "type": "hospital",
    "importance": 0.201,
    "icon": "http:\/\/nominatim.openstreetmap.org\/images\/mapicons\/health_hospital.p.20.png",
    "address": {
        "road": "Koumoundourou",
        "city": "Rethymno",
        "county": "Rethymni Municipality",
        "state_district": "Region of Crete",
        "state": "Decentralized Administration of Crete",
        "country": "Greece",
        "country_code": "gr"
    }
}]

Answer 1

尝试：

JSONArray jsonarray = new JSONArray(result);
for (int i = 0; i < jsonarray.length(); i++) {
    JSONObject jsonobject = jsonarray.getJSONObject(i);
    JSONObject hospital2 = jsonobject.getJSONObject("address")
    String hospitalName = hospital2.getString("hospital");
    String hospitalLat = hospital.getString("lat");
    String hospitalLon = hospital.getString("lon");
}

Answer 2

查看此tutorial有关如何填充ListView的信息。只需创建一个包含您正在寻找的字符串的简单类，而不是您创建的教程中的类。

关于将JSON数据解析为ListView

2 个答案: