请帮助我获得$ randomQ和$ matchinA，然后如果键入$ matchinA则刷新jQuery？

Question

Please, show me how to get a random question, '$randomQuestion'(i.e., "What color is the sky?"), and matching answer, $matchingAnswer, from the quizID section of my SQL database, and then use jQuery to refresh the form, only? I started by creating the form and some JavaScript that may work.

FORM:

<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 =

This is extremely complicated for me, and I've been having a world of trouble with it. PLEASE, comment. Thank you.

Answer 1

Well to start with, your HTML is invalid: You're not closing your tags:

<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>

As for the ajax, I would suggest using jQuery. It makes things much easier.

$.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)