在android中找不到JSON解析运行时错误源答案

【问题标题】：JSON Parsing runtime error source not found in android在android中找不到JSON解析运行时错误源
【发布时间】：2013-12-07 10:42:41
【问题描述】：

我正在尝试进行 json 解析以获取城市状态和邮政编码。我正在为此使用 google api。

我在两个设备 android 2.3 和 android 4.1.2 上调试以下代码。在 android 2.3 中它工作正常。但在 android 4.1.2 中我得到运行时错误。

这是我的代码

String url = "http://maps.google.com/maps/api/geocode/json?address="+address+"&sensor=true";
JSONParser jparser = new JSONParser();
JSONObject jobject = jparser.getJSONFromUrl(url);



try {
      DefaultHttpClient httpClient = new DefaultHttpClient();
      HttpPost httpPost= new HttpPost(url);
      HttpResponse httpResponse = httpClient.execute(httpPost);
      HttpEntity httpEntity = httpResponse.getEntity();
      is = httpEntity.getContent();
}catch (UnsupportedEncodingException e) {
      e.printStackTrace();
} catch (ClientProtocolException e) {
      e.printStackTrace();
} catch (IOException e) {
     e.printStackTrace();
}

在HttpResponse httpResponse = httpClient.execute(httpPost); 行上，我收到一个运行时错误 Source not fount from ActivityThread.performLaunchActivity。请解决我的问题

【问题讨论】：

标签： java android json google-maps

【解决方案1】：

我也面临同样的问题。这个可以帮助你。在AsyncTaskdoInBackground方法中使用下面的函数并检查响应然后你可以在onPostExecute方法中解析它

  public static String getResponefromURL(String url) {
            InputStream is = null;
            String result = "";

            // http post
            try {
                HttpClient httpclient = new DefaultHttpClient();
                HttpPost httppost = new HttpPost(url.trim());
                HttpResponse response = httpclient.execute(httppost);
                HttpEntity entity = response.getEntity();
                is = entity.getContent();

            } catch (Exception e) {
                Log.e("log_tag", "Error in http connection " + e.toString());
            }

            // convert response to string
            try {
                BufferedReader reader = new BufferedReader(new InputStreamReader(is, "iso-8859-1"), 8);
                StringBuilder sb = new StringBuilder();
                String line = null;
                while ((line = reader.readLine()) != null) {
                    sb.append(line + "\n");
                }
                is.close();
                result = sb.toString();
            } catch (Exception e) {
                Log.e("log_tag", "Error converting result " + e.toString());
            }
            return result;
        }

【讨论】：

【解决方案2】：

请参考https://stackoverflow.com/a/18259495/1944782，附加url参数并根据您的要求制作url。完成此操作后，请将您的代码放在 AsyncTask 后台并从您的主 Activity 调用 AsyncTask。希望这样做后您的问题会得到解决。

更新

请参考以下代码：

public JSONObject getAddressData() {

HttpGet httpGet = new HttpGet("http://maps.google.com/maps/api/geocode/json?address="+address+"&sensor=true");
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;

}

【讨论】：

我照你说的做了，但我还是在同一个地方遇到了同样的错误。

【解决方案3】：

我希望这对您有所帮助，您可以使用此代码获得确切的地址位置

public String gpsadd() {
    try {


    if(isGPS) {


                Geocoder gc = new Geocoder(Home.this, Locale.getDefault());
         try 
          {
            addflg=0;
            lat=location.getLatitude();
            lng=location.getLongitude();
            boolean lop=true;
            while(lop) {
            List<Address> addresses = gc.getFromLocation(lat, lng, 1);
            if (addresses.size() > 0) 
            {
                lop=false;
                addflg=1;
                address = addresses.get(0);
                s_addr=address.getAddressLine(0)+" "+address.getAddressLine(1)+" "+address.getAddressLine(2);
                disp="Location\n "+address.getAddressLine(0)+"\n"+address.getAddressLine(1)+"\n"+address.getAddressLine(2);

            }

            }


          }
          catch (Exception e)
          {
              e.printStackTrace();
          }
    } else {
        display("GPS not Enabled");

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

    }

    return disp;
}

【讨论】：