如何在PHP中更新db表的接受/拒绝

Question

我创建了一个约会表，其中user1将添加约会，user2将接受/拒绝约会。现在我的问题是当点击接受和拒绝它显示在表中但它没有在数据库中更新我已经做了很多研究但是在路障后遇到了路障，所以我真的希望你的帮助在此，非常感谢！

这是我的约会表的图像：

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

你有错误。您正在使用$_GET请求并将其设置为url。像下面一样更改你的html标签

 <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

并在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();
}

}

我希望它会对你有所帮助。如果有任何问题，请告诉我