在Json中發送查詢結果

Question

我希望以Json格式發送在我的數據庫上創建的查詢結果。查詢返回多條記錄，我希望用Json發送所有thoes記錄。我的查詢包含字段即。 Restaurant_name，Address，type of food。等等 那麼我該怎麼去？這是我不完整的代碼。

<?php 


require_once 'DB_connect.php'; 
class populatelist{ 
private $con; 
private $conn; 

function __construct() 
{ 
$this->con = new DB_connect(); 
$this->conn = $this->con->connectWithRestaurant(); 
} 
function selectallfields(){ 
$query = "SELECT * FROM `restaurant_time` LIMIT 50"; 
$query_exec = $this->conn->query($query); 
if($query_exec->num_rows >0){ 
    $queryresult = $query_exec->fetch_assoc(); 


} 
} 
} 

?> 

我這樣做的主要目的是在AndroidListView中檢索這些數據。將其填充到列表中。這裏是我在android中的代碼片段。

class Jsonfetch extends AsyncTask<String,Void,Void>{ 

    @Override 
    protected Void doInBackground(String... params) { 
     try { 
      URL url = new URL("http://172.16.16.88/orderspot/populatelist.php"); 
      HttpURLConnection httpURLConnection = (HttpURLConnection)url.openConnection(); 
      httpURLConnection.setRequestMethod("POST"); 
      httpURLConnection.setAllowUserInteraction(false); 
      InputStream inputStream = httpURLConnection.getInputStream(); 
      BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(inputStream)); 
      String temp; 
      StringBuilder stringBuilder = new StringBuilder(); 
      while((temp = bufferedReader.readLine())!=null){ 
       stringBuilder.append(temp); 
      } 
      String JsonResponse = stringBuilder.toString(); 
      try { 
       JSONArray new_array =new JSONArray(JsonResponse); 
       //int count; 
       for(int i = 0, count = new_array.length() ; i < count ; i++){ 
        JSONObject jsonObject = new_array.getJSONObject(i); 
        Resname.add(jsonObject.getString("restname")); 
        Restaddress.add(jsonObject.getString("restadd")); 
        Resttime.add(jsonObject.getString("restime")); 
        images.add(jsonObject.getInt("images")); 
       } 
      } catch (JSONException e) { 
       e.printStackTrace(); 
      } 
     } catch (MalformedURLException e) { 
      e.printStackTrace(); 
     } catch (IOException e) { 
      e.printStackTrace(); 
     } 


     return null; 
    } 
} 

Answer 1

這是更好地利用發送這樣

$query['query'] = $yourQueryResult; 
echo json_encode($query); 

查詢結果

Answer 2

希望它能幫助你。

echo json_encode($queryresult); 

編號：http://php.net/manual/en/function.json-encode.php