我有这种形式包含动态表:php MySql UPDATE不工作?
<form action="<?php echo $editFormAction;?>" method="post" name="form1">
<input type="hidden" value="<?php echo (int) $_GET['a'];?>" name="pat_id">
<table>
<tr>
<th>#</th>
<th>ORDER DESC</th>
<th>STATUS</th>
<th>SENDING DATE</th>
<th>DELIVERING NO</th>
<th>DELIVERING DATE</th>
<th>COMMENT</th>
</tr>
<?php do { ?>
<tr>
<td><?php echo $row_medi['order_no']; ?></td>
<td><?php echo $row_medi['ORDER_DESC']; ?></td>
<td><select name="mydropdown[<?php echo $row_medi['order_no']; ?>]"><option value="Undelivered">Undelivered</option><option value="Delivered">Delivered</option></select></td>
<td><?php $sqldate=date('d-m-Y',strtotime($row_medi['SENDING_DATE']));
echo $sqldate; ?></td>
<td><?php echo $row_medi['DELIVERING_NO']; ?></td>
<td><?php $sqldate=date('d-m-Y',strtotime($row_medi['DELIVERING_DATE']));
echo $sqldate; ?></td>
<td><?php echo $row_medi['COMMENT']; ?></td>
</tr>
<?php } while ($row_medi = mysql_fetch_assoc($medi)); ?>
</table>
<input type="submit" value="Save" />
<input type="hidden" name="MM_insert" value="form1">
</form>
当你看到有一个与各行的下拉菜单,我想从任何该下拉菜单的更新与选择的值孔列。没有错误消息出现,但点击保存按钮后,选定的行不会更新?这是更新代码:
$editFormAction = $_SERVER['PHP_SELF'];
if (isset($_SERVER['QUERY_STRING'])) {
$editFormAction .= "?" . htmlentities($_SERVER['QUERY_STRING']);
}
if ((isset($_POST["MM_update"])) && ($_POST["MM_update"] == "form1")) {
$pat_id = (int) $_POST['pat_id'];
$updateSQL = sprintf("UPDATE orders SET STATUS=%s, pat_id=%s WHERE order_no=%s",
GetSQLValueString($_POST['mydropdown'], "text"),
GetSQLValueString($_POST['pat_id'], "int"),
GetSQLValueString($_POST['order_no'], "int"));
mysql_select_db($database_PPS, $PPS);
$Result1 = mysql_query($updateSQL, $PPS) or die(mysql_error());
$updateGoTo = "Pharmacy.php?a=$pat_id";
if (isset($_SERVER['QUERY_STRING'])) {
$updateGoTo .= (strpos($updateGoTo, '?')) ? "&" : "?";
$updateGoTo .= $_SERVER['QUERY_STRING'];
}
header(sprintf("Location: %s", $updateGoTo));
}
你是什么意思“任何这种下降的更新与选择的值孔列下拉菜单'?每行的选择名称相同,因此$ _REQUEST ['mydropdown']是下拉列表的最后一行值? – regilero
当我尝试echo($ choice)时,所选选项的值也不会显示!我如何解决这个问题? – 7kemZmani