[英]ajax post to php file
我无法使其正常工作-这是我的首次尝试。 我确实意识到该信息尚未序列化,我只是想让它首先工作。
有任何想法我在这里做错了吗? 它不会将信息保存到mysql数据库中。
index.html:
<head>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js"></script>
<script type="text/javascript" src="purchase.js"></script>
</head>
<form name="paypal_form" onsubmit="return validate_form();" method="post">
<input type="text" size="25" name="os0" value="">
<input type="text" size="25" name="os1" value="">
<input type="hidden" name="item_name" value="product">
<input type="image" src="images/paypal.gif" name="submit">
</form>
Payment.js:
function validate_form()
{
// code to validate form data
// ....
valid = true;
var regcode = document.paypal_form.os0.value;
var email = document.paypal_form.os1.value;
var product = document.paypal_form.item_name.value;
var dataString = 'regcode=' + regcode + '&email=' + email + '&product=' + product;
$.ajax({
url: "/process.php",
type: "POST",
data: dataString,
success: function()
{
alert("Order Submitted");
}
});
return valid;
}
process.php:
$dbhost = "localhost";
$dbuser = "dbuser";
$dbpass = "dbpass";
$dbname = "dbname";
$date = date('Y/m/d');
$RegCode = $_POST['regcode'];
$Email = $_POST['email'];
$Product = $_POST['product'];
mysql_connect($dbhost, $dbuser, $dbpass);
// Store the transaction ID in the database
mysql_query("INSERT into payment (date, regcode, sentemail, status) values ('$date', '$RegCode', '$Email', '$Product')");
如果process.php在同一个文件夹中,则将其设为url:'process.php',然后
mysql_connect($dbhost, $dbuser, $dbpass);
mysql_select_db("dbname", $con);
在表单标签中添加以下属性,然后尝试
action="javascript:void(0);"
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