PHP 表列

Question

所以我的 PHP 显示了一个 MySQL 表，我知道你可以按我的计划对这些进行分类。但如下图所示，添加按钮添加了另一列。

是否可以使两列下方的添加按钮更新和删除？如果可以，我该怎么做。这是我的代码，对于任何错误或困惑，我深表歉意，我还在学习。

while ($record = mysql_fetch_array($myData)) {
  echo "<form action=usermanagement.php method=post>";
  echo "<tr>";
  echo "<td>" . "<input type=text name=id value=" . $record['ID'] . " </td>";
  echo "<td>" . "<input type=text name=rank value=" . $record['Rank'] . " </td>";
  echo "<td>" . "<input type=text name=username value=" . $record['Username'] . " </td>";
  echo "<td>" . "<input type=text name=password value=" . $record['Password'] . " </td>";
  echo "<td>" . "<input type=hidden name=hidden value=" . $record['ID'] . " </td>";
  echo "<td>" . "<input type=submit name=update value=update" . " </td>";
  echo "<td>" . "<input type=submit name=delete value=delete" . " </td>";
  echo "</tr>";
  echo "</form>";
}
echo "<form action=usermanagement.php method=post>";
echo "<tr>";
echo "<td><input type=text name=uid></td>";
echo "<td><input type=text name=urank></td>";
echo "<td><input type=text name=uusername></td>";
echo "<td><input type=text name=upassword></td>";
echo "<td>" . "<input type=submit name=add value=add" . " </td>";
echo "</form>";
echo "</table>";    

Answer 1

注意输入的结束标记。隐藏的输入可以放在表格中的任何位置，因为它们不会显示:

while($record = mysql_fetch_array($myData)) {
    echo '<form action=usermanagement.php method=post>
              <tr>
                   <td>
                        <input type=text name=id value="' . $record['ID'] . '" />
                   </td>
                   <td>
                        <input type=text name=rank value="' . $record['Rank'] . '" />
                   </td>
                   <td>
                        <input type=text name=username value="' . $record['Username'] . '" />
                   </td>
                   <td>
                        <input type=text name=password value="' . $record['Password'] . '" />
                   </td>
                   <td>
                        <input type=hidden name=hidden value="' . $record['ID'] . '" />
                        <input type=submit name=update value=update />
                   </td>
                   <td>
                        <input type=submit name=delete value=delete />
                   </td>
              </tr>
          </form>';
}

echo '<form action=usermanagement.php method=post>
          <tr>
               <td>
                    <input type=text name=uid />
               </td>
               <td>
                    <input type=text name=urank />
               </td>
               <td>
                    <input type=text name=uusername />
               </td>
               <td>
                    <input type=text name=upassword />
               </td>
               <td colspan="2">
                    <input type=submit name=add value=add />
               </td>
          </tr>
      </form>';

此外，尝试停止使用 mysql_* 函数，因为它们已被弃用。请改用 PDO、mysqli_ 或 MySQLi 类。