PHP

Question

我可以使我的数据的编码json不是这个

的格式

{"tbl_accepted":[{"renterLat":"15.706376","renterLng":"121.065852"}]}

我只想得到Lat和Lng的值，所以我可以相应地在单独的textViews中显示它们。这就是我如何得到它。

private void getJSON(String url){
    class GetJSON extends AsyncTask<String, Void, String>{
        protected String doInBackground(String... params){
            String uri = params[0];
            BufferedReader bufferedReader = null;
            try{

                URL url = new URL(uri);
                HttpURLConnection con = (HttpURLConnection) url.openConnection();
                StringBuilder sb = new StringBuilder();
                bufferedReader = new BufferedReader(new InputStreamReader(con.getInputStream()));

                String json;
                while((json=bufferedReader.readLine()) != null){
                    sb.append(json+"\n");
                }
                return sb.toString().trim();
            }catch(Exception e){
                return null;
            }
        }

        @Override
        protected void onPostExecute(String s){
            super.onPostExecute(s);
            tvLat.setText(s);
            Log.d(TAG,tvLat.getText().toString());
        }
    }
    GetJSON gj = new GetJSON();
    gj.execute(url);
}

PHP

<?php
include_once("connection.php");
$where = '';
if (isset($_GET['renter'])){
$where = " WHERE renter like '%".addslashes($_GET['renter'])."%'";
}
$sql = "SELECT renterLat, renterLng FROM tbl_accepted ".$where." ORDER BY acceptedID DESC";

$result = mysqli_query($conn, $sql);
$json = array();

if(mysqli_num_rows($result)){

while($row = mysqli_fetch_assoc($result)){

    $json['tbl_accepted'][]=$row;
}
}

mysqli_close($conn);
echo json_encode($json);

?>

先谢谢先生。

Answer 1

您在onPostExecute中收到了关注字符串。

{"tbl_accepted":[{"renterLat":"15.706376","renterLng":"121.065852"}]}

使用JSONObject和JSONArray将 JSON字符串转换为对象。

要将字符串转换为对象，您可以在onPostExecute中看起来像这样。

@Override
        protected void onPostExecute(String s){
            super.onPostExecute(s);
            JSONObject latObj=new JSONObject(s);
            JSONArray tbl_accepted=latObj.getJSONArray('tbl_accepted');
            tvLat.setText(tbl_accepted.getJSONObject(0).getString('renterLat'));
            Log.d(TAG,tvLat.getText().toString());
        }

Android Studio从数据库获取数据到TextView

PHP

1 个答案: