随机生成的问题在android应用程序中是行不通的

Question

answered_questions:

​

​

问题：

​

​

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);

}

在我用android运行这个应用程序之前，一切都正常。当我单击getQuestion按钮/调用该方法时，应用程序将生成已回答的问题。该应用程序的目的是生成一个问题，这个问题不存在于“回答的问题”表中，只需单击一次

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);
}

Answer 1

为什么不简单地合并两个表并添加一个布尔字段(tinyint) answered？然后，您只需将SQL请求更改为SELECT question FROM questions WHERE category = '$category' AND answered = 0即可。