php - 请帮我获取 $randomQ 和 $matchinA，然后在输入 $matchinA 时刷新 jQuery？

Question

请告诉我如何从 quizID 部分获取随机问题“$randomQuestion”(即“天空是什么颜色？”)和匹配答案 $matchingAnswer我的SQL数据库，然后使用jQuery来刷新表单，仅此而已？我首先创建了表单和一些可能有效的 JavaScript。

表格:

<form name="$quizID" action="http://asite.com" method="post">
  <fieldset>
    <legend="$randomQuestion">
    <p>
      <label>Answer: <input type="text" id="answer" onkeydown="submitAns(submit.id)" /></label>
    </p>
  </fieldset>
</form>

JS:

function submitAns(id) {
  if (document.getElementById(id).value=="$matchingAnswer")
          document.a.submit();
}

JQUERY:

$.post('get-question_matchinganswer_for_quizID.php', {
  quizID: $quizID,
  question },
  function(data) {
    alert('Question is: ' + data.question)
    alert('Answer is: ' + data.answer)
  },
  'json'
);

PHP:

<!-- Help -->
function random_row($table, $column) {
  $max_sql = "SELECT max(" . $column . ") 
          AS max_id
          FROM " . $table;
  $max_row = mysql_fetch_array(mysql_query($max_sql));
  $random_number = mt_rand(1, $max_row['max_id']);
  $random_sql = "SELECT * FROM " . $table . "
          WHERE " . $column . " >= " . $random_number . " 
          ORDER BY " . $column . " ASC
          LIMIT 1";
  $random_row = mysql_fetch_row(mysql_query($random_sql));
  if (!is_array($random_row)) {
    $random_sql = "SELECT * FROM " . $table . "
            WHERE " . $column . " < " . $random_number . " 
            ORDER BY " . $column . " DESC
            LIMIT 1";

    $random_row = mysql_fetch_row(mysql_query($random_sql));
  }
  return $random_row;
}

$randomQuestion =
$matchinAnswer =

这对我来说非常复杂，我一直在为此烦恼。请给出意见。谢谢。

Answer 1

首先，您的 HTML 无效:您没有关闭标签:

<form name="$quizID" action="http://asite.com" method="post">
    <fieldset>
        <legend><?php echo $randomQuestion?></legend>
        <label>
            Answer: <input type="text" id="answer"
                           onkeydown="submitAns(submit.id)" />
        </label>
    </fieldset>
</form>

至于ajax，我建议使用jQuery。它使事情变得更加容易。

$.post('getqidqandanswer.php', {
    quizID: 1337,
    questionID: 42},
    function(data) {
        alert('Question is: ' + data.question)
        alert('Answer is: ' + data.answer)
    },
    'json'
);

PHP:

$quizID = isset($_POST['quizID']) ? $_POST['quizID'] : null
$questionID = isset($_POST['questionID']) ? $_POST['questionID'] : null

if($quizID && $questionID)
{
   $data = getQuestionData($quizID, $questionID)
}
elseif($quizID)
{
   $data = getRandomQuestionData($quizID)
}
else
{
    $data = array(
        'question' => '',
        'answer'   => ''
    )
}

echo json_encode($data)