我需要使用php更改具有不同id的表的html元素

我有两个不同id的表，我希望表1显示db中的食物项目，表2显示衣服。然而，我不知道如何使用php访问和更改id。我知道ajax和jquery会使它更简单，但我只想学习如何在后端实现这一点

<h3>Lunch/drinks </h3>
<table id= '1' border="1">
<tr>
  <td>Product Name</td>
  <td>Product price</td>
  <td>Product image</td>
</tr>

<h4> Uniforms </h4>
<table id = '2' border="1">
<tr>
  <td>Product Name</td>
  <td>Product price</td>
  <td>Product image</td>
</tr>

<?php
include("dbconnect.php");
$sql = "Select PRODUCTNAME, PRODUCTPRICE From PRODUCT Where PRODUCTType = 'food'";
$query = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_array($query)) {
  echo
   "<tr>
    <td>{$row['PRODUCTNAME']}</td>
    <td>{$row['PRODUCTPRICE']}</td>
       </tr>\n";
}
$sqlq = "Select PRODUCTNAME, PRODUCTPRICE From PRODUCT Where PRODUCTType = 'Uniform'";
$queryq = mysqli_query($conn, $sqlq);
while ($row = mysqli_fetch_array($queryq)) {
  echo
   "<tr>
    <td>{$row['PRODUCTNAME']}</td>
    <td>{$row['PRODUCTPRICE']}</td>
       </tr>\n";
}
?>
</table>
</table>

午餐/饮料
品名
产品价格
产品形象
制服
品名
产品价格
产品形象
只要重新安排你已经拥有的东西。你就快到了：

<?php
include("dbconnect.php");
?>
<h3>Lunch/drinks </h3>
<table id= '1' border="1">
<tr>
  <td>Product Name</td>
  <td>Product price</td>
  <td>Product image</td>
</tr>
<?php
$sql = "Select PRODUCTNAME, PRODUCTPRICE From PRODUCT Where PRODUCTType = 'food'";
$query = mysqli_query($conn, $sql);
while ($row = mysqli_fetch_array($query)) {
  echo
   "<tr>
    <td>{$row['PRODUCTNAME']}</td>
    <td>{$row['PRODUCTPRICE']}</td>
   </tr>\n";
}
?>
</table>

<h4> Uniforms </h4>
<table id = '2' border="1">
<tr>
  <td>Product Name</td>
  <td>Product price</td>
  <td>Product image</td>
</tr>

<?php
$sqlq = "Select PRODUCTNAME, PRODUCTPRICE From PRODUCT Where PRODUCTType = 'Uniform'";
$queryq = mysqli_query($conn, $sqlq);
while ($row = mysqli_fetch_array($queryq)) {
  echo
   "<tr>
    <td>{$row['PRODUCTNAME']}</td>
    <td>{$row['PRODUCTPRICE']}</td>
    </tr>\n";
}
?>
</table>


午餐/饮料
品名
产品价格
产品形象