Question

我正在尝试使用PHP更新数据库中的表。

我使用循环来获取表中的所有数据。然后，添加＆＃34;批准＆＃34;用户可以单击的每行前面的按钮，如果他们想要批准请求，则会将状态更改为“已批准”。

现在我被卡住了，我无法找到更新状态字段的方法，而且我对JavaScript不满意。

我想让按钮名称动态，但我不能 Output image

这是循环代码：

while($row = mysqli_fetch_array($result))
  {

echo "<tr bgcolor='#cccccc'>";
echo "<td>" . $row['nursename'] . "</td>";
echo "<td>" . $row['dr'] . "</td>";
echo "<td>" . $row['natureofreq'] . "</td>";
echo "<td>" . $row['clinic'] . "</td>";
echo "<td>" . $row['clinicfloor'] . "</td>";
echo "<td>" . $row['qitem'] . "</td>";
echo "<td>" . $row['moreinfo'] . "</td>";
echo "<td>" . $row['user_check'] . "</td>";
echo "<td>" . $row['id'] . "</td>";
echo "<td bgcolor='#009FE3'>" . $row['rstatus'] . "</td>";
echo '<form name="submit" method="post" action="">';
echo "<td>" . '<button id="approve" name="approve"> Approve</button>' . "</td>";
echo '</form>';
echo "</tr>";
}

这是更新代码，无效：

$id=$row['id'];
if(isset($_POST['approve']))
{
  $allowed = mysqli_query($n," UPDATE newrequest SET rstatus = 'Approved' WHERE id = '$id' ");
}

我将不胜感激。

output

Answer 1

     $url_of_current_page = $_SERVER['REQUEST_URI'];

     if(isset($url_of_current_page)){

         $_SESSION['url_of_current_page'] = $url_of_current_page;

     }

     $approve_row = '';

     while($row = mysqli_fetch_array($result)){

         $approve_row .= "<tr bgcolor='#cccccc'>

                             <td>" . $row['nursename'] . "</td>

                             <td>" . $row['dr'] . "</td>

                             <td>" . $row['natureofreq'] . "</td>

                             <td>" . $row['clinic'] . "</td>

                             <td>" . $row['clinicfloor'] . "</td>

                             <td>" . $row['qitem'] . "</td>

                             <td>" . $row['moreinfo'] . "</td>

                             <td>" . $row['user_check'] . "</td>

                             <td bgcolor='#009FE3'>" . $row['rstatus'] . "</td>

                             <td>
                                 <a href='action.php?id=" . $row['id'] . "'>
                                     <button>Approve</button>
                                 </a>
                             </td>

                         </tr>";

    }


     if(isset($approve_row)){

          echo $approve_row;

     }

动作代码。如果您的操作页面不是action.php，请告诉我页面的名称。

     if(isset($_GET['id'])){

         $id = $_GET['id'];

         mysqli_query($n," UPDATE newrequest SET rstatus = 'Approved' WHERE id = '$id' ");

         if(isset($_SESSION['url_of_current_page'])){

              header( "Location: ". $_SESSION['url_of_current_page'] );
              exit();

         }
     }

Answer 2

您似乎没有从您丢弃的代码

向数据库发送任何数据

使用PHP，HTML和MySQL更新表

2 个答案: