我有下拉选择,当我单击提交按钮时,它会根据下拉列表的值显示记录。它工作正常,但是当我刷新 app_list2.php 时它什么也没显示。我想保留 app_list2.php 中显示的记录。如何将所有 $_POST 值存储在 session 中?
<?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>";
}
?>