表单回应PHP不使用ajax提交

我有一个从数据库中回显的表单,但问题是当我尝试提交时,只有第一个回显表单提交,其余表示不提交.以下是我的代码.

editquestion.phh

<thead>
          <tr>
            <th style="width: 5%;">S/N</th>
            <th style="width: 20%;">QUESTION</th>
            <th style="width: 40%;">ANSWER</th>
            <th style="width: 30%;">KEYWORDS</th>
            <th style="width: 5%;">SAVE/UPDATE</th>
            </tr>
      </thead>
      <tbody>
        <?PHP
        $sql = $db->prepare("SELECT * FROM questions");
        $result = $sql->execute();

        while ($row = $result->fetchArray(sqlite3_ASSOC))
        {
            $quiz_id = $row['quiz_id'];
            $question = $row['question'];
            $answer = $row['answer'];
            $keywords = $row['keywords'];

            echo '<form action="updatequestion.PHP" method="post" enctype="multipart/form-data">
            <tr>
            <td><input style="width: 50%" type="text" name="cid" id="cid" value="'.$quiz_id.'"></td>
            <td><input type="text" name="question" id="question" value="'.$question.'"></td>
            <td><input type="text" name="answer" id="answer" value="'.$answer.'"></td>
            <td><input type="text" name="keywords" id="keywords" value="'.$keywords.'"></td>
            <td><input type="submit" name="qupdate" class="qupdate" value="Update"></td>
            </tr>
            </form>';

        }
        ?>
        </tbody>
  </table>

qupdate.js

$(document).ready(function() {
$('.qupdate').click(function() {
    question = $('#question').val();
    answer = $('#answer').val();
    keywords = $('#keywords').val();
    id = $('#cid').val();

    $.ajax({
        type: "POST",
        url: "updatequestion.PHP",
        data: "cid="+id+"&question="+question+"&answer="+answer+"&keywords="+keywords,
        success: function(html){
            if(html = "true")
            {
                $('.qupdate').css("opacity", "1");
            }
            else
            {
                alert("not successful");
            }
        },
        beforeSend: function(){
            $('.qupdate').css("opacity", "0.5");
        }
    });
    return false;
});
});

刚刚添加了updatequestion.PHP的代码.

<?PHP
session_start();
require_once("db.PHP");
$db = new MyDB();


if (isset($_POST['question']) || isset($_POST['answer']) || isset($_POST['cid']))
{
$id = strip_tags(@$_POST['cid']);
$cname = strip_tags(@$_POST['question']);
$cunit = strip_tags(@$_POST['answer']);
$keywords = strip_tags(@$_POST['keywords']);

if (empty($cname) || empty($cunit))
{
    echo "fill";
}
else
{
    $sql = $db->prepare("UPDATE questions SET question = ?, answer = ?, keywords = ? WHERE quiz_id = ?");
    $sql->bindParam(1, $cname, sqlite3_TEXT);
    $sql->bindParam(2, $cunit, sqlite3_TEXT);
    $sql->bindParam(3, $keywords, sqlite3_TEXT);
    $sql->bindParam(4, $id, sqlite3_INTEGER);

    $result = $sql->execute();

    if ($result)
    {
        echo "true";
    }
    else
    {
        echo "false";
    }
}
}
?>

但是ajax似乎只适用于第一个回显的数据,似乎没有提交其余的数据.我该如何解决这个问题？

提前致谢.

解决方法:

Add class dynamic-form to form tag and remove id from all fields:

echo '<form class="dynamic-form"  action="updatequestion.PHP" method="post" enctype="multipart/form-data">
            <tr>
            <td><input style="width: 50%" type="text" name="cid" value="'.$quiz_id.'"></td>
            <td><input type="text" name="question" value="'.$question.'"></td>
            <td><input type="text" name="answer" value="'.$answer.'"></td>
            <td><input type="text" name="keywords" value="'.$keywords.'"></td>
            <td><input type="submit" name="qupdate" class="qupdate" value="Update"></td>
            </tr>
            </form>';

Update in JS

$(document).ready(function () {
    $('.dynamic-form').on('submit', function () {
        var formdata = $(this).serialize();
        $.ajax({
            type: "POST",
            url: "updatequestion.PHP",
            data: formdata,
            success: function (html) {
                //success
            }
        });
        return false;
    });
});