将JSON表单错误响应转换为html字符串

时间:2014-05-31 21:24:33

标签: jquery ajax json

看到我无法找到解决问题的方法,我可能会错误地解决这个问题。

我尝试使用AJAX构建表单,该表单将在提交表单时提供错误响应或成功响应。我能够收到发送到我的表单的错误响应,但它是在JSON中,这显然看起来很糟糕。目前我正在使用简化形式,只有两个要求:(1)名称必须是John和(2)日期必须是将来。

这是我的HTML:

<!DOCTYPE html>
<head>
    <title>AJAX Form</title>
</head>
<body>

    <form action="ajax/contact.php" method="post" class="ajax">
        <div>
            <input type="text" name="name" placeholder="Your name">
        </div>
        <div>
            <input type="text" name="email" placeholder="Your email">
        </div>
        <div>
            <input type="text" placeholder="" class="textinput" name="departuredate" id="departing">
        </div>
        <div>
        <textarea name="message" placeholder="Your message"></textarea>
        </div>
        <input type="submit" value="Send">
    </form>

    <div id="ack"></div>

    <script type="text/javascript" src="js/jquery-1.11.0.js"></script>
    <script type="text/javascript" src="js/main.js"></script>
    <script type="text/javascript" src="js/jquery-ui-1.10.4.custom.js"></script>

    <script>
    $(document).ready(function() {

        $("#departing").datepicker({
            //minDate: 0,
            //maxDate: "+1y",
            dateFormat: "yy-mm-dd",
            defaultDate: new Date(),
        });
    });
    </script>
</body>
</html>

这是我的PHP文件:

<?php

$errors = array();
$form_data = array();

$name = htmlspecialchars($_POST['name']);
$email = htmlspecialchars($_POST['email']);
$date = htmlspecialchars($_POST['departuredate']);
$message = htmlspecialchars($_POST['message']);

if (date('Y-m-d') > $date) {
$errors['date'] = "Date is in the past";
goto end;
}
if ($name != 'John') {
$errors['name'] = "Name is not John";
goto end;
}

end:

if (empty($errors)) {
$form_data['success'] = true;
} else {
$form_data['errors'] = $errors;
}
echo json_encode($form_data);
?>

这是我的AJAX文件:

$('form.ajax').on('submit', function() {
var that = $(this),
    url = that.attr('action'),
    type = that.attr('method');

$.ajax({
    url: url,
    type: type,
    data: that.serialize(),
    dataType: 'json',
    cache: false,
    success: 
        function(result) {
        if(!result.success) {
            $('#ack').html(JSON.stringify(result.errors));
            console.log(result.errors);
        } else {
        console.log('Valid date and name entries');
        }
    }
});

return false;
});

谢谢!

0 个答案:

没有答案