Classes for view file?

I am not sure how to explain in detail but I will try my best.

Currenly I have a php file it look something like this:

   $SQL_Cat = mysql_query("SELECT * FROM categories WHERE restaurant_id = '$restaurantID'");
    while ($category = mysql_fetch_array($SQL_Cat))
    {
        $CategoryID = $category['id'];
        echo $category['name'];

        $q = mysql_query("SELECT * FROM items WHERE category_id = '" . $CategoryID . "'");
        while($item = mysql_fetch_array($q))
        {          
            $item_name = $item['name'];
            echo $item_name;                
             }
        }

As you can see there are $category and $item ,how to create this class (classname/{Category, Item}) to use the view. Example below:

<?php foreach($categories as $category): ?>
    <table  border=0 Cellspacing='0'>
        <tr>
            <td>
              <?php echo $category->name; ?>
            </td>
        </tr>
    <?php foreach ($category->items as $item): ?>       
        <tr>
            <td>
              <?php echo $item->name; ?>
            </td>
 开发者_StackOverflow社区       </tr>
    <?php endforeach; ?>
    </table>
<?php endforeach; ?>

So I can keep SQL queries away from a view file.

$SQL_Cat = mysql_query("SELECT * FROM categories WHERE restaurant_id = '$restaurantID'");
while ($category = mysql_fetch_object($SQL_Cat))
{
    $category->items = array();
    $q = mysql_query("SELECT * FROM items WHERE category_id = '" . $category->id . "'");
    while($item = mysql_fetch_object($q))
    {          
        $category->items[] = $item;
    }

    $categories[] = $category;
}

Use PDO and fetch it into an object. PDO is capable of automatically constructing objects so you don't have to write class definitions.