單查詢，以檢查結果

Question

我有這個表（演示）：

questions 
---------- 
question_id | question  | choice1 | choice2 | choice3 | choice3 | choice4 | answer 
--------------------------------------------------------------------------------------- 
1   | what is..  | apple | orange | grapes | pinea | blah | 2 
2   | what is..  | apple | orange | grapes | pinea | blah | 4 
3   | what is..  | apple | orange | grapes | pinea | blah | 1 
4   | what is..  | apple | orange | grapes | pinea | blah | 2 
5   | what is..  | apple | orange | grapes | pinea | blah | 3 
6   | what is..  | apple | orange | grapes | pinea | blah | 4 

一組5個隨機問題將被顯示給用戶，然後用戶選擇的答案。所以，當用戶提交表單（答案）時，如何檢查用戶提交的是否正確？

我現在要做的是，通過在提交answersheet每個question_id循環，然後做一個查詢來檢查，如果答案是正確的這樣的（僞碼）：

$wrong_questions = $total_questions_in_answersheet; 
foreach($question in $answersheet) 
{ 
    $c = SELECT COUNT(*) FROM `questions` WHERE `question_id` = $question_id AND `answer` = $answer; 

    if($c > 0) 
    $wrong_questions--; 
} 

if($wrong_questions > 0) 
    echo 'FAILED'; 
else 
    echo 'WELL DONE'; 

是否有可能做在單個查詢中檢查答案？如果是這樣，那麼最好這樣做嗎？

Answer 1

$where = implode(' OR ', array_map(function($question) { 
    return "(question_id = $question[question_id] AND answer != $question[answer])"; 
}, $answersheet)); 

$sql = "SELECT COUNT(*) AS wrong_questions FROM questions WHERE $where"; 
$result = mysqli_query($con, $sql); 
$row = mysqli_fetch_assoc($result); 
if ($row['wrong_questions'] > 0) { 
    echo 'FAILED'; 
} else { 
    echo 'WELL DONE'; 
} 

Answer 2

在循環中運行查詢通常被認爲是不好的做法;所以，其實我只會做兩件事： 首先，SELECT所有正確答案掛在id，在關聯數組中，如$answers['id']['answer']; 二，比較提交的答案與$answers數組id。