更新时无法发布值？

Question

代码：

<?php
if(isset($_POST['save']))
  {
    $comment1 = $_POST['comment2'].",".date('Y-m-d');
    $comment2 = $_POST['comment2'];
    $date = date('Y-m-d');
    $chk = implode(",", $_POST['chk']);
    $id = explode(",", $chk);
    foreach ($id as $id) 
    {
      echo "update enquires2 set comment1='$comment1' , comment2='$comment2' , s_date='$date' where id='$id'";
    } 
  }
?>
<form name="myform" method="POST">
  <table>
    <tr>
      <th>History</th>
      <th>Add Comment</th>
      <th>Action</th>
    </tr>

    <?php
      $sql = "select * from enquires2";
      $result = mysqli_query($link,$sql);
      while ($row = mysqli_fetch_array($result)) 
      {
          echo "<tr>
                  <td><input type='checkbox' name='chk[]' value=".$row['id']."></td>
                  <td>".$row['comment1']."</td>
                  <td>
                    <input type='text' name='comment2' id='comment2' />
                  </td>
                </tr>";
      }
      echo "<tr>
              <td>
                <input type='submit' name='save' value='save'>
              </td>
            </tr>";
    ?>
  </table>
</form>

在此代码中，当我在comment2中插入值时，选中复选框并单击“保存”按钮，然后打印查询，如图所示。它没有发表评论2的价值。我该如何解决这个问题？请帮忙。

谢谢