[英]JavaScript not updating form action
我没有附上CSS,因为这个问题无关紧要,但是由于某种原因我的表单无法提交,我知道您必须指定一个操作来执行此操作,这是我在javascript中所做的,但是每当我单击按钮时添加到购物车只需重新加载此页产品 。 已经尝试将addtoCart.php放在操作中,但是这会导致其他错误不会像我需要的那样带出产品ID。 那么,JavaScript是否有任何理由不更新我的表单操作?
<body class="oneColFixCtrHdr">
<div id="container">
<form name="myform" id="myform" action="" method="post"/>
<input type="hidden" name="PHPSESSID" value="<?php echo session_id; ?>">
<input type="hidden" name="cartNumber" value="<?php echo $cartNumber ?>">
<input type="hidden" name="productID" id="productID" value="">
<div id="header">
<table width="760">
<tr>
<td width="188" rowspan="2">
<a href="default.php"> <img src="images/CongaMoe.jpg" width="150" height="153" alt="Conga Moe Logo" border="0"/></a>
</td>
<td width="361"><img src="images/homeTitle.jpg" width="343" height="152"></td>
<td width="188">
<a href="default.php"> <img src="images/CongaMoe.jpg" width="150" height="153" alt="Conga Moe Logo" border="0"></a>
</td>
</tr>
<tr>
<td colspan="2" align="right">
<a href="drums.php"><img src="images/btnCongaDrums.jpg" width="100" height="32" border="0"></a> <a href="products.php"><img src="images/btnBuyAConga.jpg" width="100" height="32" border="0"></a> <a href="viewcart.php"><img src="images/btnLookInCart.jpg" width="100" height="32" border="0"></a> <a href="checkout.php"><img src="images/btnPayAndGo.jpg" width="100" height="32" border="0"></a>
</td><td width="3"></td>
</table>
</div>
<!-- end #header -->
<div id="mainContent">
<script type="text/javascript">
function addToCart(product, formObj) {
document.getElementById("productID").value = product;
document.getElementById("myform").action = "addtoCart.php";
formObj.submit;
}
</script>
<table>
<?php
require_once('appVars.php');
$dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME)
or die('Could not connect to MySQL server as CIS55Student, cis55 database.');
$strQuote = chr(34);
$query = "SELECT * FROM products_nalanirowe";
$rs = mysqli_query($dbc, $query);
$num_rows = mysqli_num_rows($rs);
for ($i = 0; $i < $num_rows; $i++) {
$row = mysqli_fetch_array($rs);
$productID = $row['productID'];
$productName = $row['productName'];
$productPrice = $row['productPrice'];
echo "<tr class='productText'>";
echo "<td valign='top'><input type='image' name='submit' src='images/btnAddToCart.jpg' width='100' height='32' border='0' onclick'javascript:addToCart(". $strQuote . "$productID" . $strQuote . ",this.form)' ></td>";
echo "<td valign='top'>". $row['productID'] . "</td>";
echo "<td valign='top'>". $row['productName'] . "</td>";
echo "<td valign='top'". $row['productDesc'] . "</td>";
echo "<td valign='top'>". $row['productPrice'] . "</td>";
echo "<td valign='top'><img src='" . IMAGEPATH . $row['productImgName'] . "' border='0'></td>";
echo "</tr>";
}
mysqli_close($dbc);
?>
</table>
<!-- end #mainContent --></div>
<div id="footer">
<p><strong>©Conga Moes 2010</strong></p>
<!-- end #footer --></div>
</form>
<!-- end #container --></div>
</body>
onclick'javascript:addToCart(". $strQuote . "$productID" . $strQuote . ",this.form)'
,您没有正确分配处理程序,并且此formObj.submit;
应该是formObj.submit();
另一个错误是
onclick'javascript:addToCart(". $strQuote . "$productID" . $strQuote . ",this.form)' ,
应该读
onclick='javascript:addToCart(". $strQuote . "$productID" . $strQuote . ",this.form)' ,
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