地理coder.getFromLocation函数抛出＆QUOT;超时等待服务器响应＆QUOT;例外 [英] geocoder.getFromLocation function throws "Timed out waiting for server response" exception

问题描述

我想从MainActivity开始IntentService来获得用户的位置。里面的服务，我设法扭转地理$ C C A try块内的位置$但是当我捕获该异常并打印它说：超时等待服务器响应exception.But几次，我已经拿到了location.so我觉得有什么错我的code.But如果抛出异常8超时10.So，你可以建议一些事情，以避免这种情况就不会是有益的。

I am trying to get location of user by starting IntentService from MainActivity. Inside the service i try to reverse geocode the location inside a try block but when i catch the exception and print it says "Timed out waiting for server response" exception.But a few times I have got the location.so i think there is nothing wrong with my code.But it won't be useful if it throws exception 8 times out of 10.So can you suggest some thing to avoid this.

推荐答案

下面是一个简单的工作液：

Here is a simple and working solution:

public static JSONObject getLocationInfo(String address) {
    StringBuilder stringBuilder = new StringBuilder();
    try {

    address = address.replaceAll(" ","%20");    

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


        response = client.execute(httppost);
        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) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }

    return jsonObject;
}
private static List<Address> getAddrByWeb(JSONObject jsonObject){
    List<Address> res = new ArrayList<Address>();
    try
    {
        JSONArray array = (JSONArray) jsonObject.get("results");
        for (int i = 0; i < array.length(); i++)
        {
            Double lon = new Double(0);
            Double lat = new Double(0);
            String name = "";
            try
            {
                lon = array.getJSONObject(i).getJSONObject("geometry").getJSONObject("location").getDouble("lng");

                lat = array.getJSONObject(i).getJSONObject("geometry").getJSONObject("location").getDouble("lat");
                name = array.getJSONObject(i).getString("formatted_address");
                Address addr = new Address(Locale.getDefault());
                addr.setLatitude(lat);
                addr.setLongitude(lon);
                addr.setAddressLine(0, name != null ? name : "");
                res.add(addr);
            }
            catch (JSONException e)
            {
                e.printStackTrace();

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

    }

    return res;
}

现在只需更换

geocoder.getFromLocation(locationAddress, 1); 

与

getAddrByWeb(getLocationInfo(locationAddress));  

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