javascript - 如何在 PHP 中更新数据库表的接受/拒绝

Question

我创建了一个约会表，其中 user1 将添加约会，而 user2 将接受/拒绝约会。 现在我的问题是当点击接受和拒绝时它显示在表格中但它没有在数据库中更新我对此做了很多研究但是遇到了一个又一个障碍, 所以我真的希望你能在这方面提供帮助，非常感谢!

这是我的约会表的图片:

这是我的代码:

       <table cellpadding="0" cellspacing="0" border="0" class="table table-condensed" id="example"> 
                    <thead>
                          <tr>
                                <th>appoinment ID</th>
                              <th>Date</th>
                                <th>time</th>
                              <th>teacher</th>
                             <th>parent</th>
                              <th> accept/reject </th>
                              <th>state</th>
                            </tr>
                   </thead>
                   <tbody>
                        <?php 
                            $query=mysqli_query($conn, "select * from `app` left join `par` on par.par_id=app.par_id
                            left join `tea` on tea.tea_id=app.tea_id
                            ORDER BY app_id DESC");

                                if($query === false)
                                {
                                    throw new Exception(mysqli_error($conn));
                                }
                                while($row=mysqli_fetch_array($query))
                                {
                                    $ann_id=$row['app_id'];
                                    $date=$row['date'];
                                    $msg=$row['time'];

                                    $username = $row['username'];
                                     $username = $row['p_username'];
                        ?>
                <tr>
                    <td><?php echo $row['app_id'] ?></td>
                     <td> <?php echo date('j/m/y',strtotime($row['date'])); ?></td>
                    <td><?php echo $row['time'] ?></td>
                   <td><?php echo $row['p_username'] ?></td>
                <td><?php echo $row['username'] ?></td>

                    <td>
        <a href="#" class="reject">reject</a> 
        <a href="#" class="accept">accept</a>
    </td>
    <td>
        <div class="chgtext">PENDING</div>
    </td>

                </tr>

                        <?php   

    if (isset($_GET['state'], $_GET['app_id'])) {
$stmt = mysqli_prepare($conn, "UPDATE app SET state = ? WHERE app_id = ?");
mysqli_stmt_bind_param($stmt, "sd", $_GET['state'], $_GET['s.app_id']);
$stmt->execute();
$stmt->close();
}

                                 }

                       ?>

                  </tbody>
              </table>
         </div>
      </form>

Answer 1

你那里有一个错误。您正在使用 $_GET 请求并在 url 中进行设置。像下面这样更改您的 html 标记

 <a href="tea_appview.php?app_id=<?php echo $row['app_id'] . "&" . "state=reject";?>" class="reject">reject</a>

<a href="tea_appview.php?app_id=<?php echo $row['app_id'] . "&" . "state=accept";?>" class="reject">accept</a> <!-- I assumed that column names in database were app_id and state

在 php 代码中

if (isset($_GET['state'], $_GET['app_id'])) {
$stmt = mysqli_prepare($conn, "UPDATE app SET state = ? WHERE app_id = ?");
mysqli_stmt_bind_param($stmt, "sd", $_GET['state'], $_GET['app_id']);
$stmt->execute();
$stmt->close();
}

}

希望对您有所帮助。如果有任何问题，请告诉我