2017-09-01 52 views
0

answered_questions:随机生成的问题不会在Android应用工作

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问题:

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我有这些表和下面的PHP脚本

if($_SERVER['REQUEST_METHOD'] == 'GET') { 


//makes it work 
$category = (string)filter_input(INPUT_GET, 'category'); 
$game_id = (string)filter_input(INPUT_GET, 'game_id'); 

require_once('dbConnect.php'); 

$query = "SELECT question FROM questions 
    WHERE category = '$category' 
    and question 
    NOT IN 
    (SELECT question 
    FROM answered_questions 
    WHERE game_id='$game_id')ORDER BY Rand() limit 1"; 

$r = (mysqli_query($con, $query)); 

$res = mysqli_fetch_array($r); 

$result = array(); 

array_push($result, array(
     "question" => $res['question'], 
    ) 
); 

echo json_encode(array("result" => $result)); 

mysqli_close($con); 

} 

一切工作,直到我在android中运行应用程序。当我点击getQuestion按钮/调用方法时,该应用程序产生已经回答的问题。该应用旨在产生一个问题,是不是在与每个回答问题表点击

private void getQuestion() { 

    String url =""; 

    String cat = category.getText().toString(); 
    String id = game_id.getText().toString(); 

    if (cat.equals("Control Questions")){ 
     url = "http://192.168.0.20/Articulate/getControlQuestion.php?game_id="+id; 
    }else { 
     url = "http://192.168.0.20/Articulate/getQuestion.php?category="+cat+"&game_id="+id; 
    } 

    StringRequest stringRequest = new StringRequest(url, new Response.Listener<String>() { 
     @Override 
     public void onResponse(String response) { 
      showJSON(response); 
     } 
    }, 
      new Response.ErrorListener() { 
       @Override 
       public void onErrorResponse(VolleyError error) { 
       } 
      }); 

    RequestQueue requestQueue = Volley.newRequestQueue(this); 
    requestQueue.add(stringRequest); 
} 

private void showJSON(String response){ 
    String ques=""; 
    try { 
     JSONObject jsonObject = new JSONObject(response); 
     JSONArray result = jsonObject.getJSONArray(Config.JSON_ARRAY); 
     JSONObject collegeData = result.getJSONObject(0); 
     ques = collegeData.getString("question"); 
    } catch (JSONException e) { 
     e.printStackTrace(); 
    } 
    question.setText(ques); 
} 

回答

0

你们为什么不干脆合并两个表,并添加一个布尔(TINYINT)字段answered?那么你只需要将你的SQL请求更改为SELECT question FROM questions WHERE category = '$category' AND answered = 0

+0

状态可以被回答,空或传递,并且需要我改变应用程序的其余部分。而这仍然可能无法正常工作,考虑到上述查询在应用程序 – cmc12345

+0

中的应用程序将由许多用户播放时似乎不工作,所以如果有一个表,则表示没有game_id,表会被更新对于所有用户 – cmc12345

+0

好吧好吧^^,我试着创建你的数据库并执行你写的请求,它看起来没问题。我认为问题在其他地方,也许是放入网址 – RasAlGhul