發送參數，PHP和獲取引起JSON

Question

我有發送參數代碼PHP和獲取引起JSON：

private void details_location(int id){ 

     HashMap<String, String> parameter = new HashMap<>(); 
     parameter.put("id", id); 

     JsonObjectRequest jsonObjReq = new JsonObjectRequest(Method.GET, url, new JSONObject(parameter), new Response.Listener<JSONObject>() { 
      @Override 
      public void onResponse(JSONObject response) { 
       Log.d(TAG, response.toString()); 

       try { 
        JSONArray respon=(JSONArray)response.get("location"); 

        for (int i = 0; i < respon.length(); i++) { 
         JSONObject person = (JSONObject) respon.get(i); 
         id_location= person.getString("id_location"); 
         address= person.getString("address");     
        } 
       } catch (JSONException e) { 
        e.printStackTrace(); 
        Toast.makeText(getApplicationContext(), 
          "Error: " + e.getMessage(), 
          Toast.LENGTH_LONG).show(); 
       } 
      } 
     }, new Response.ErrorListener() { 

      @Override 
      public void onErrorResponse(VolleyError error) { 
       VolleyLog.d(TAG, "Error: " + error.getMessage()); 
       Toast.makeText(getApplicationContext(), 
         error.getMessage(), Toast.LENGTH_SHORT).show(); 
      } 
     }); 
    } 

，我有文件PHP對於等待處理的查詢MySQL和具有輸出JSON：

<?php 
    require_once('db_connect.php'); 
    if(isset($_GET['id'])){ 
     $id=$_GET['id']; 
     $action="SELECT * FROM locaions WHERE id_location=\"$id\";"; 
     $query=mysqli_query($db_connect, $action); 

     if (mysqli_num_rows($query) > 0) 
     { 
      $json['location']=array(); 
      while($row=mysqli_fetch_assoc($query)){ 
       $data=array(); 
       $data["id_location"]=$row["id_location"]; 
       $data["address"]=$row["address"]; 

       array_push($json['location'], $data); 
      } 
     } 
     mysqli_close($db_connect); 
     echo json_encode($json); 
    } 
    else{ 
     $action="SELECT * FROM locaions WHERE id_location=4;"; 
     $query=mysqli_query($db_connect, $action); 

     if (mysqli_num_rows($query) > 0) 
     { 
      $json['location']=array(); 
      while($row=mysqli_fetch_assoc($query)){ 
       $data=array(); 
       $data["id_location"]=$row["id_location"]; 
       $data["address"]=$row["address"]; 

       array_push($json['location'], $data); 
      } 
     } 
     mysqli_close($db_connect); 
     echo json_encode($json); 
    } 
?> 

當我運行該程序時，結果總是顯示位置id_location=4（語句else）。爲什麼isset($_GET['id']) = false？如何將參數發送到PHP並獲得JSON結果？

Answer 1

這對於發送參數烏爾PHP服務器

protected String doInBackground(String... params) { 
      // TODO Auto-generated method stub 

      try { 
       String parameter1 = (String) params[0]; 
       String parameter2 = (String) params[1]; 

       String link = "Mention ur url link here/parameter1+parameter1 +parameter2+parameter2+"; 

       URL url = new URL(link); 
       HttpClient client = new DefaultHttpClient(); 
       HttpGet request = new HttpGet(); 
       request.setURI(new URI(link)); 
       HttpResponse response = client.execute(request); 
       BufferedReader in = new BufferedReader(new InputStreamReader(
         response.getEntity().getContent())); 

       StringBuffer sb = new StringBuffer(""); 
       String line = ""; 

       while ((line = in.readLine()) != null) { 
        sb.append(line); 
        break; 
       } 
       in.close(); 
       return sb.toString(); 
      } 

      catch (Exception e) { 
       return new String("Exception: " + e.getMessage()); 
      } 
     } 

在後期執行PHP應該在JSON格式返回數據，並可以處理這個數據在後期執行的代碼