这是我的jquery移动应用程序的代码。但是我用php
结果创建的表单不起作用。表单不会在提交时发布数据。我甚至试过data-ajax="flase"
。
<table id="stock">
<? $sql = mysqli_query($con, "SELECT * FROM products WHERE `product_id` = '1'");
while($row=mysqli_fetch_array($sql)){
$name = $row['name'];
$chamecal = $row['chamecal'];
$price = $row['selling_price'];
$bprice = $row['buying_price'];
$quantity = $row['quantity'];
$manufacturer = $row['manufacturer'];
$date = $row['date'];
$edate = $row['expire'];
$pid = $row['product_id'];
?>
<form id="stockupdate" method="post" action="medupdate.php">
<tr>
<td><?=$name;?></td>
<td><?=$chamecal;?></td>
<td><?=$manufacturer;?></td>
<td><input type="text" name="medbprice" value="<?=$bprice;?>" /></td>
<td><input type="text" name="medprice" value="<?=$price;?>" /></td>
<td><?=$date;?></td>
<td><?=$edate;?></td>
<td><input type="text" name="medstock" value="<?=$quantity;?>" /></td>
<td class="ui-screen-hidden"><input type="text" name="pid" value="<?=$pid;?>" /></td>
<td>
<input type="submit" title="Update" data-icon="gear" data-theme="f" data-inline="true" data-iconpos="notext" />
</td>
</tr>
<?
}
?>
</form>
</table>
值处理程序文件medupdate.php
<?php session_start();?>
<?php include_once('include/config.php');?>
<?
$pid = $_POST['pid'];
$medbprice = $_POST['medbprice'];
$medprice = $_POST['medprice'];
$medstock = $_POST['medstock'];
$sql = mysqli_query($con, "UPDATE `products` SET `quantity` = '$medstock', `buying_price` = '$medbprice' , `selling_price` = '$medprice' WHERE `product_id` = '$pid'");
if($sql){
echo 1;
}
?>
答案 0 :(得分:1)
如果我冒险猜测,那是因为你正在放置
<form id="stockupdate" method="post" action="medupdate.php">
直接在一个表元素中(这是无效的标记)。将它移到桌子外面(在<table>
之前)。当您创建无效标记时,浏览器引擎会尝试更正它,这可能会导致此类问题。
这就是chrome对你的标记做的事情:
答案 1 :(得分:1)
您无法关注html结构,form
内的table
,
<table>
<form>
<tr>
<td>
</td>
</tr>
</form>
</table>
这是工作代码,
表单php
<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="width=device-width, initial-scale=1">
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
<link rel="stylesheet" href="http://code.jquery.com/mobile/1.3.2/jquery.mobile-1.3.2.min.css">
<script src="http://code.jquery.com/jquery-1.9.1.min.js"></script>
<script src="http://code.jquery.com/mobile/1.3.2/jquery.mobile-1.3.2.min.js"></script>
</head>
<body>
<form id="stockupdate" method="post" action="medupdate.php" data-ajax="false">
<table>
<tr>
<td><input type="text" name="medbprice" value="xxxx" /></td>
<td><input type="text" name="medprice" value="yyyyy" /></td>
<td><input type="text" name="medstock" value="zzzzzz" /></td>
<td class="ui-screen-hidden"><input type="text" name="pid" value="111111" /></td>
<td>
<input type="submit" title="Update" data-icon="gear" data-theme="f" data-inline="true" data-iconpos="notext" />
</td>
</tr>
</table>
</form>
</body>
</html>
<强> medupdate.php 强>
<?php
session_start();
$pid = $_POST['pid'];
$medbprice = $_POST['medbprice'];
$medprice = $_POST['medprice'];
$medstock = $_POST['medstock'];
echo $pid;
// your database operations
?>