3

好的，所以我查看了论坛并看到了与此类似的问题，但是我仍然无法使用从各个页面窃取的代码。

我正在使用引导 css 来显示联系表格、常用字段、姓名电子邮件、消息等。我将用户添加到 mysql 数据库中的 action="process.php" 然后通过电子邮件向我发送确认有人已提交表格。所以在这方面一切都很好，只是我想在提交表单后显示一个简单的“谢谢”消息，而不是通过重定向到另一个页面。

我有以下信息：

<!-- thank you message -->
<div id="thanks" class="alert alert-success fade">
  <button href="#" type="button" class="close">&times;</button>
  <h4>Got it!</h4>
  <p>Thanks. I've received your message, I'll be in touch within 24 hours. Promise.</p>
</div>

然后使用这个js添加“in”以在提交后显示它：

$('#send_button').click(function () {
    $('#thanks').addClass('in'); 
});

$('.close').click(function () {
    $(this).parent().removeClass('in'); // hides alert with Bootstrap CSS3 implem
});

我简要地看到了感谢警报消息，但随后我被重定向到“process.php”，这没有显示任何内容，因为其中没有 html，只有 mysql 的东西和 php 邮件程序。还有一点需要注意，这是否感兴趣，我最初通过 ajax 加载联系表单，所以 url 就像 wdr/index.php#contact

有人可以帮我完成代码。我敢肯定，我缺少一些简单的东西来使这项工作正常进行。

任何帮助表示赞赏。

科尔

4

1 回答 1

4

使用 Ajax 很容易。这是我用于简单发送的内容：

js：

$('#form_id').on('submit', function(e) {
    e.preventDefault(); //Prevents default submit
    var form = $(this); 
    var post_url = form.attr('action'); 
    var post_data = form.serialize(); //Serialized the form data for process.php
    $('#loader', form).html('<img src="../img/forms/loader.gif" /> Please Wait...');
    $.ajax({
        type: 'POST',
        url: 'process.php', // Your form script
        data: post_data,
        success: function(msg) {
            $(form).fadeOut(500, function(){
                form.html(msg).fadeIn();
            });
        }
    });
});

进程.php：

<?php

/* Configuration */
$subject = 'Submission received'; // Set email subject line here
$mailto  = 'your email address'; // Email address to send form submission to
/* END Configuration */

$firstName      = $_POST['firstName'];
$lastName       = $_POST['lastName'];
$email          = $_POST['email'];
$companyName    = $_POST['companyName'];
$phone          = $_POST['phone'];
$callTime       = $_POST['callTime'];
$timestamp = date("F jS Y, h:iA.", time());

// HTML for email to send submission details
$body = "
<br>
<p>The following information was submitted through the contact form on your website:</p>
<p><b>Name</b>: $firstName $lastName<br>
<b>Email</b>: $email<br>
<b>Company name</b>: $companyName<br>
<b>Phone number</b>: $phone (Please call in the <b>$callTime</b>)</p>
<p>This form was submitted on <b>$timestamp</b></p>
";

// Success Message
$success = "
<div class=\"row-fluid\">
    <div class=\"span12\">
        <h3>Submission successful</h3>
        <p>Thank you for taking the time to contact Pacific One Lending. A representative will be in contact with you shortly. If you need immediate assistance or would like to speak to someone now, please feel free to contact us directly at <strong>(619) 890-3605</strong>.</p>
    </div>
</div>
";

$headers = "From: $firstName $lastName <$email> \r\n";
$headers .= "Reply-To: $email \r\n";
$headers .= "MIME-Version: 1.0\r\n";
$headers .= "Content-Type: text/html; charset=ISO-8859-1\r\n";
$message = "<html><body>$body</body></html>";

if (mail($mailto, $subject, $message, $headers)) {
    echo "$success"; // success
} else {
    echo 'Form submission failed. Please try again...'; // failure
}

?>
于 2013-08-11T17:55:35.733 回答