我有一个像这样的ajax弹出窗口......
<div class="white-popup-block" style="max-width:600px; margin: 20px auto;">
<?php
include "konekdb.php";
if(isset($_GET['kode_kb'])) {
$kode_kb = mysql_real_escape_string($_GET['kode_kb']);
$sql = "SELECT * FROM kb_tiny
WHERE kode_kb='$kode_kb'" ;
$query = mysql_query($sql, $koneksiDB) or die ("error query".mysql_error());
$data = mysql_fetch_assoc($query);
echo "Title : ".$data['title'];
echo "No Ticket :".$data['no_ticket'];
echo "Error Code : ".$data['error_code'];
echo "example : ".$data['ex_case'];
echo "Solution :".$data['solution'];
echo "Creator : ".$data['creator'];
echo "Date Modified : ".$data['modified'];
}
?>
</br>
</br>
</br>
<form method="POST" action="<?php echo $_SERVER['PHP_SELF']; ?>">
<input name="approved" type="button" value="Approved" /> <input name="reject" type="button" value="Reject" />
</form>
</div>
当我在我的选择 * 数据库上单击详细信息时,它会显示类似上面的弹出窗口,当它弹出它提供按钮以批准和拒绝时...如何处理该按钮,以便它可以更新我的数据库状态...像这样:
if(isset($_POST['approved'])) {
$sql = "UPDATE `kb_tiny` SET `status` = 'approved'";
请帮助我如何假设代码工作......数据已经显示但批准按钮并拒绝不起作用:(