Question

方法Geocoder.getFromLocationName()会在Android 4.1上引发异常Service not available，即使GooglePlayServicesUtil.isGooglePlayServicesAvailable()返回SUCCESS而Geocoder.isPresent()返回true。
新的Google Maps v2 API中是否有任何地理编码的官方示例？

Answer 1

Geocoder与Google Maps Android API v2无关。

您可能希望直接使用Google Geocoding API而不是Geocoder，这会为您提供有限的数据量，并可能受设备或Android版本特定问题的影响。

Answer 2

试试这个......

      public  JSONObject getLocationFormGoogle(String placesName) {

    HttpGet httpGet = new HttpGet("http://maps.google.com/maps/api/geocode/json?address=" +placesName+"&ka&sensor=false");
    HttpClient client = new DefaultHttpClient();
    HttpResponse response;
    StringBuilder stringBuilder = new StringBuilder();

    try {
        response = client.execute(httpGet);
        HttpEntity entity = response.getEntity();
        InputStream stream = entity.getContent();
        int b;
        while ((b = stream.read()) != -1) {
            stringBuilder.append((char) b);
        }
    } catch (ClientProtocolException e) {
    } catch (IOException e) {
    }

    JSONObject jsonObject = new JSONObject();
    try {
        jsonObject = new JSONObject(stringBuilder.toString());
    } catch (JSONException e) {

        e.printStackTrace();
    }

    return jsonObject;
}

public  LatLng getLatLng(JSONObject jsonObject) {

    Double lon = new Double(0);
    Double lat = new Double(0);

    try {

        lon = ((JSONArray)jsonObject.get("results")).getJSONObject(0)
            .getJSONObject("geometry").getJSONObject("location")
            .getDouble("lng");

        lat = ((JSONArray)jsonObject.get("results")).getJSONObject(0)
            .getJSONObject("geometry").getJSONObject("location")
            .getDouble("lat");

    } catch (JSONException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

    return new LatLng(lat,lon);

}



LatLng Source =getLatLng(getLocationFormGoogle(placesName));

Geocoder.getFromLocationName（）在Android 4上使用Google Maps V2引发“服务不可用”例外

2 个答案: