无法从 nominatim url (openstreetmap) 得到 JSON

Question

我无法从反向地理编码的 URL JSON 响应中获取结果。我也收到错误：

"W/System.err: org.json.JSONException: End of input at character 0 of "

这里是urlhttps://nominatim.openstreetmap.org/reverse?format=geojson&lat=14.6458&lon=121.0949

HttpDataHandler.java

public String GetHTTPData(String requestUrl)
{
    URL url;
    String response = "";
    try{
        url = new URL(requestUrl);
        HttpURLConnection conn = (HttpURLConnection)url.openConnection();
        conn.setReadTimeout(15000);
        conn.setConnectTimeout(15000);
        conn.setRequestMethod("GET");
        conn.setRequestProperty("Content-Type","application/x-www-form-urlencoded");
        conn.setDoOutput(true);
        int responseCode = conn.getResponseCode();

        if(responseCode == HttpURLConnection.HTTP_OK)
        {
            String line;
            BufferedReader br = new BufferedReader(new InputStreamReader(conn.getInputStream()));
            while((line = br.readLine()) != null)
                response+=line;
        }
        else
            response = "";

MainActivity

 @Override
    protected String doInBackground(String... strings) {
        try{
            double lat = Double.parseDouble(strings[0].split(",")[0]);
            double lng = Double.parseDouble(strings[0].split(",")[1]);
            String response;
            HttpDataHandler http = new HttpDataHandler();
            //String url = String.format("http://open.mapquestapi.com/geocoding/v1/reverse?key=2KtQuwfGGdfHxj6ybdgqcC7uFHrgVoJy&location=%.4f,%.4f",lat,lng);
            String url = String.format("https://nominatim.openstreetmap.org/reverse?format=json&lat=%.4f&lon=%.4f",lat,lng);
            response = http.GetHTTPData(url);
            Log.d("testpandebug,doinbg", response);
            Log.d("testpandebug,doinbg", url);
            return response;
        }
        catch (Exception ex)
        {

        }
        return null;
    }

我正在使用日志从 url 获取响应。使用其他地理编码器 URL（注释字符串 URL）时，它 return 是一个响应，但是，使用 nominatim 不会 return 响应。我想使用 nominatim，因为它 return 是一种更准确的反向地理编码。

Answer 1

如果它 return 不是有效的 body，则查看 HTTP 响应 header。我猜你违反了 Nominatim's usage policy。您提供有效的 HTTP 用户代理吗？

无法从 nominatim url (openstreetmap) 得到 JSON

Can't get JSON from nominatim url (openstreetmap)

java

android

json

openstreetmap

nominatim