answered_questions:
的问题:
我有这些表和以下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);
}
一切正常,直到我在android中运行应用程序。当我单击getQuestion按钮/调用方法时,应用程序会生成已经回答的问题。该应用程序旨在生成一个问题,每次点击都不在已回答的问题表中
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);
}
答案 0 :(得分:0)
为什么不简单地合并两个表并添加一个布尔(tinyint)字段answered
?然后你只需要将你的SQL请求更改为SELECT question FROM questions WHERE category = '$category' AND answered = 0