刷新后如何保留在同一选择中？
[英]How to retain in the same selection after refresh?
问题描述
How to retain with the same selection in dropdown when refresh or after I submit or delete? 
刷新时或提交或删除后，如何在下拉菜单中保留相同的选择？
I have this code below that select value from database. 我下面的代码从数据库中选择值。 I want to back with the same value in select dropdown after I refresh or I submit the form or delete records. 刷新或提交表单或删除记录后，我想在选择下拉列表中使用相同的值。
bid.php bid.php
<html>
<head>
<script>
function showUser(str) {
var $txtHint = $('#txtHint');
if (str == "") {
$txtHint.html('');
return;
}
$txtHint.load('bid_list.php?q=' + str)
}
</script>
</head>
<body onload=showUser(str="ALL")>
<?php
$mysqli = new mysqli("localhost", "root", "", "app");
$result = $mysqli->query("SELECT bid FROM procurement WHERE bid LIKE '13-___' OR bid LIKE '1_-___' OR bid LIKE '2_-___' GROUP BY bid ORDER BY bid");
$option = '';
while($row = $result->fetch_assoc())
{
$option .= '<option value = "'.$row['bid'].'">'.$row['bid'].'</option>';
}
?>
<select name="users" onchange="showUser(this.value)" style="overflow:scroll;width:100px;">
<option value="ALL" selected='ALL'>ALL</option>
<?php echo $option; ?>
</select>
<div id="txtHint">
</div>
</body>
</html>
bid_list.php bid_list.php
<?php
$mysqli = new mysqli("localhost", "root", "", "app");
$q = $_GET["q"];
$where = '';
if ( $q != 'ALL' ) {
$where = " WHERE bid='$q' ";
}
$result1 = $mysqli->query("
SELECT bid, item_name, item_description, unit, unit_cost, quantity, supplier, po_number, po_date, counter, SUM(unit_cost*quantity) AS total_amount
FROM procurement
$where
GROUP BY counter ORDER BY bid
");
echo'<table id="tfhover" cellspacing="0" class="tablesorter" style="text-transform:uppercase;">
<thead>
<tr>
<th colspan="3" id="none" style="cursor: default;"></th>
<th title="Item Name">Item Name</th>
<th title="Item Description">Description</th>
<th title="Example : Pc, Pcs, Box and Etc.">Unit</th>
<th title="Item Price">Unit Cost</th>
<th title="Total Item Quantity">QTY</th>
<th title="Total Price">Total Amount</th>
<th title="Name of Supplier">Supplier</th>
<th title="Purchase Order Date">PO Date</th>
<th title="Purchase Order #">PO #</th>
<th id="none" style="cursor: default;"></th>
</tr>
</thead>';
echo'<tbody>';
while($row = $result1->fetch_assoc()){
if($row['bid'] != '')
{
echo'<tr>
<td align="center"><a href="bid.php?de='.$row["counter"].'" onclick="return confirm(\'Really want to delete ?\');"><img src="images/del.png" border="0" width="10" height="10" title="Delete"></a></td>
<td align="center"><a class="fancybox" href="edit.php?pn='.$row["counter"].'"><img src="images/edit.png" border="0" width="10" height="10" title="Edit"></a></td>
<td align="center"><a class="fancybox" href="comments.php?pn='.$row["counter"].'"><img src="images/remarks.png" border="0" width="10" height="10" title="Remarks and Notes"></a></td>
<td>'.$row['item_name'].'</td>
<td>'.$row['item_description'].'</td>
<td>'.$row['unit'].'</td>
<td>'.number_format($row['unit_cost'], 2, '.', ',').'</td>
<td>'.$row['quantity'].'</td>
<td>'.number_format($row['total_amount'], 2, '.', ',').'</td>
<td>'.$row['supplier'].'</td>
<td>'.$row['po_date'].'</td>
<td>'.$row['po_number'].'</td>
<td><input name="checkbox[]" type="checkbox" id="checkbox[]" value="'.$row['counter'].'"></td>
</tr>';
}
}
echo "</tbody></table>";
2 个解决方案
解决方案1
1 2013-10-09 00:43:47
use a jquery ajax call, and update the DOM when your ajax call succesfully returns. 使用jquery ajax调用，并在ajax调用成功返回时更新DOM。 see http://api.jquery.com/jQuery.ajax/ 参见http://api.jquery.com/jQuery.ajax/
解决方案2
1 2013-10-09 01:05:16
AJAX can be used for interactive communication with a database, please check the web site here : http://www.w3schools.com/php/php_ajax_database.asp AJAX可用于与数据库进行交互通信，请在此处查看网站： http : //www.w3schools.com/php/php_ajax_database.asp
i believe that is what you need. 我相信那是您所需要的。
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