所以幾個小時以來我一直在努力解決這個問題。任何幫助將不勝感激 Homepage- index.php連接器.php<?php$connect = mysqli_connect("localhost", "dummy", "123456", "www") or die ('I cannot connect to the database  because: ' . mysql_error())?>插入.php<?php  include 'conn.php'; // MySQL Connectionif(!empty($_POST))  {  $output = '';  $message = '';  $ostatus = mysqli_real_escape_string($connect, $_POST["ostatus"]);  if($_POST["order_oid"] != '')  {  $query = "  UPDATE ordersSET ostatus='$ostatus',   WHERE oid='".$_POST["order_oid"]."'";  $message = 'Data Updated';  }  else  {  $query = "  INSERT INTO orders(ostatus)  VALUES('$ostatus');  ";  $message = 'Added Record Successfully';  }  if(mysqli_query($connect, $query))  {  $output .= '<label class="text-success">' . $message . '</label>';  $select_query = "SELECT * FROM orders ORDER BY id DESC";  $result = mysqli_query($connect, $select_query);  $output .= '  <table class="table table-bordered">  <tr>  <th width="70%">Customer Name</th><th width="70%">Customer Address</th><th width="70%">Customer Mobile</th><th width="70%">Payment Method</th><th width="70%">Order Status</th><th width="15%">Edit</th>  <th width="15%">View</th>  </tr>  ';  while($row = mysqli_fetch_array($result))  {  $output .= '  <tr>  <td><?php echo $row["oname"]; ?></td>  <td><?php echo $row["odeladd"]; ?></td><td><?php echo $row["omobile"]; ?></td><td><?php echo $row["opaymethod"]; ?></td><td><?php echo $row["ostatus"]; ?></td><td><input type="button" name="edit" value="Edit" id="'.$row["oid"] .'" class="btn btn-info btn-xs edit_data" /></td>  <td><input type="button" name="view" value="view" id="' . $row["id"] . '" class="btn btn-info btn-xs view_data" /></td>  </tr>  ';  }  $output .= '</table>';  }  echo $output;  }  ?>有兩個名為 orders 和 ordersdata 的表，它們都有 oid 作為公共列。兩者都在同一個數據庫中，稱為 www。當我點擊編輯按鈕時，應該會顯示 ajax 彈出窗口，但事實并非如此?，F在我從編輯中獲取 oid，從視圖中獲取 id。為我已通過 order_oid 的 edit_data 調用插入頁面，為我已通過 order_id 的 view_data 調用插入頁面。幫助將不勝感激 謝謝。