以表单形式在ajax和php中填充三重下拉菜单 - populate triple drop down in ajax and php in a form

populate triple drop down in ajax and php in a form

我已经设法用php和ajax填充了我的三个下拉列表。这是我代码中显示三个下拉列表的一部分：

<?php
echo "<font id='"categoria'"><select>'n";
echo "<option value='0'>Select the Firm</option> 'n" ;
echo "</select></font>'n";
?>
<?php
echo "<font id='"subcategoria'"><select>'n";
echo "<option value='0'>Select the Claims Hub</option> 'n" ;
echo "</select></font>'n";
?>
<?php
echo "<font id='"subcategoria2'"><select>'n";
echo "<option value='0'>Select the Area</option> 'n" ;
echo "</select></font>'n"; 
?>

我的问题是：我在这些页面上还有一个表单，我想把三个下拉列表作为我表单的一部分，以便将值发布到另一个mysql表中。有什么建议我该怎么做吗？这是我的表单部分：

<form method="post" align= "right">
<table >
    <tr><td>&nbsp;</td>
    <tr>
        <td>Select the Matter Type: </td>
        <td><?php echo $dropdown; ?></td>
    </tr>
    <tr><td>&nbsp;</td>
    <tr>
         <td style="text-align: left; ">Active  </td>
         <td align="left" >
            <input type="checkbox" name="active" value="1" />
         </td>
    </tr>   
    <tr><td>&nbsp;</td>
    <tr>
        <td>&nbsp;</td>
        <td><input type="submit" name="submit" value="Add" align= "right" /></td>
    </tr>
</table>

为什么不直接将它们插入到html中呢？

<form method="post" align= "right">
<table >
<tr><td>&nbsp;</td>
<tr>
    <td>Select the Matter Type: </td>
    <td><?php echo $dropdown; ?></td>
</tr>
<tr>
  <td>
    <?php echo "<font id='"categoria'"><select><option value='0'>Select the Firm</option></select></font>";?>
  </td>
</tr>
<tr><td>&nbsp;</td>
<tr>
     <td style="text-align: left; ">Active  </td>
     <td align="left" >
        <input type="checkbox" name="active" value="1" />
     </td>
</tr>   
<tr><td>&nbsp;</td>
<tr>
    <td>&nbsp;</td>
    <td><input type="submit" name="submit" value="Add" align= "right" /></td>
</tr>