php - $_POST 変数全体をセッションに格納する方法

Question

ドロップダウンを選択して、送信ボタンをクリックすると、ドロップダウンの値に基づいてレコードが表示されます。正常に動作しますが、app_list2.php を更新すると何も表示されません。app_list2.php に表示されるレコードを保持したい。すべての $_POST 値を session に保存するにはどうすればよいですか?

    <?php
    $mysqli = new mysqli("localhost", "root", "", "app");
    if(isset($_POST['submit'])){
    $drop = $_POST['drop_1'];
    $tier_two = $_POST['tier_two'];

    $where = "WHERE app_cn='$drop' AND app_plan_no='$tier_two'";

$result1 = $mysqli->query("
    SELECT *, SUM(unit_cost*quantity) AS total_amount, SUM(unit_cost*1st_quarter) AS 1st_quarter_total, SUM(unit_cost*2nd_quarter) AS 2nd_quarter_total, SUM(unit_cost*3rd_quarter) AS 3rd_quarter_total,
    SUM(unit_cost*4th_quarter) AS 4th_quarter_total, SUM((quantity)-(1st_q_qty_del+2nd_q_qty_del+3rd_q_qty_del+4th_q_qty_del)) AS balance
    FROM app 
    $where 
    GROUP BY counter ORDER BY counter
");
?>

<?php
echo'<table id="tfhover" cellspacing="0" class="tablesorter" style="text-transform:uppercase;">
        <thead>
            <tr>
        <th>Item</th>
        <th>Description</th>
        <th>Fund Source</th>
        <th>Unit</th>
        <th>Unit Cost</th>
    </tr>
        </thead>';
        echo'<tbody>';
$i=1;
while($row = $result1->fetch_assoc()){
if($row['app_cn'] != '')
 {
 echo'<tr>
    <td>'.$row['item_name'].'</td>
        <td>'.$row['item_description'].'</td>
    <td>'.$row['fund_source'].'</td>
        <td>'.$row['unit'].'</td>
        <td>'.number_format($row['unit_cost'], 2, '.', ',').'</td>
       </tr>';
    }
        }
    echo "</tbody></table>";
}
?>