提交表单时保持表单窗口打开

Question

当用户点击链接时，请考虑此图片是iframe窗口。

我的问题

当用户点击存款时，表单会被提交并 窗口关闭 ，因此用户不知道存款是否成功。

我想做什么

我正在寻找一种方法，在提交表单后保持iframe窗口打开，以显示相应的消息

表单HTML

   <form name="depForm" action="" id="register_form" method="post">
            User Name<br /><input type="text" name="uname" value="" /><br />
            Credit Card Nr<br /><input type="text" name="cc" value="" /><br />
            CSV Nr<br /><input type="text" name="csv" value="" /><br />
            Amount<br /> <input type="text" name="amount" value="" /><br />
            <input type="submit" value="deposit" name="deposit" class="buttono" />
            </form>

PHP代码

if(isset($_POST['deposit'])){
            if(isset($_SESSION['FBID'])){
                $uid=$_SESSION['FBID'];
                $amount = $_POST['amount'];
                $cc = $_POST['cc'];
                $csv = $_POST['csv'];
                //new bal
                $bal = getBal($uid);
                $newBal = $bal+$amount;
                $sql="UPDATE members SET balance='$newBal' WHERE member_nr='$uid'";
                $result = mysql_query($sql) or die("error please try again");   
                    if($result){

                    }
            }

如果有人可以告诉我如何在表单提交后保持iframe打开，我们将不胜感激。

Answer 1

您需要更改表单提交以使用AJAX。在响应中，您可以包含状态标志，以向UI指示请求是否成功并采取适当的行动。像这样：

$('#register_form').submit(function(e) {
    e.preventDefault(); // stop the standard form submission
    $.ajax({
        url: this.action,
        type: this.method,
        data: $(this).serialize(),
        success: function(data) {
            if (data.success) {
                // show success message in UI and/or hide modal
            } else {
                // it didn't work
            }
        },
        error: function(xhr, status, error) {
            // something went wrong with the server. diagnose with the above properties
        }
    });
});

$success = false;
if (isset($_POST['deposit'])) {
    if (isset($_SESSION['FBID'])) {
        $uid = $_SESSION['FBID'];
        $amount = $_POST['amount'];
        $cc = $_POST['cc'];
        $csv = $_POST['csv'];

        $bal = getBal($uid);
        $newBal = $bal + $amount;
        $sql = "UPDATE members SET balance='$newBal' WHERE member_nr='$uid'";
        $result = mysql_query($sql) or die("error please try again");   
        if ($result) {
            $success = true;
        }
    }
}

echo json_encode(array('success' => $success));