2013-02-12 156 views
1

我非常感谢我的PHP联系表单的一些帮助。我已经遵循了一些教程来尝试使这个工作,我仍然没有运气。代码如下。PHP表单无法正常工作

N.B.我正在使用Twitter Bootstrap Framework中定义的样式元素,并且窗体出现在模式样式的弹出窗口中(尽管我看不出为什么会影响窗体)。

除了启用PHP之外,还有其他任何我可能需要在主机端激活/配置的东西。

表单代码:

<form method="POST" action="mail.php"> 
    <fieldset> 
     <input type="text" name="first_name" placeholder="First Name"> 
     <input type="text" name="last_name" placeholder="Last Name"> 
     <input type="email" name="email" placeholder="Email Address"> 
     <div class="controls controls-row"> 
     <input type="date" name="check_in" placeholder="Check in Date"><span class="help-inline">Check-in</span> 
     </div> 
     <div class="controls controls-row"> 
     <input type="date" name="check_out" placeholder="Check out Date"><span class="help-inline">Check-out</span> 
     </div> 
     <input type="number" name="rooms" min="1" max="7" placeholder="Number of Rooms"> 
     <input type="number" name="occupants" min="1" max="14" placeholder="Number of Occupants"> 
     <textarea rows="3" name="additional" input class="input-xparge" placeholder="Additional requirements" class="span5"></textarea> 
    </fieldset> 

    <div class="modal-footer"> 
     <button type="submit" input type="submit" id="submit" class="btn btn-block btn-large btn-success" value="submit">Submit</button> 
    </div> 
    </form> 

PHP代码:

<?php 
$first_name = $_POST['first_name']; 
$last_name = $_POST['last_name']; 
$email = $_POST['email']; 
$check_in = $POST['check_in']; 
$check_out = $POST['check_out']; 
$rooms = $POST['rooms']; 
$occupants = $POST['occupants']; 
$additional = $_POST['additional']; 
$from = "From: $first_name"; 
$to = "[email protected]"; 
$subject = "New Booking Enquiry"; 

$body = "First Name: $first_name\n Last Name: $last_name\n Email: $email\n Check In: $check_in\n Check Out: $check_out\n Number of Rooms: $rooms\n Number of Occupants: $occupants\n Additional Information: $additional"; 

if ($_POST['submit']) { 
if (mail ($to, $subject, $body, $from)) { 
    echo '<p>Your message has been sent!</p>'; 
    } else { 
     echo '<p>Something went wrong, go back and try again!</p>'; 
    } 
} 

>

衷心感谢你提前!

+0

你得到的错误/问题是什么? – Laurence 2013-02-12 16:53:28

+0

你应该描述什么不起作用。 – 2013-02-12 16:54:14

回答

3

你不需要添加name =“submit”到'button'标签吗?

在你的PHP

,你正在测试 '如果($ _ POST [' 提交 ']){'

和$ _POST [ '提交']可能没有得到设定没有名字在标签= “提交”

+0

这应该解决它。我想补充说,你应该过滤/清理你的用户输入。垃圾邮件发送者可以使用您的脚本以这种方式从服务器发送垃圾邮件。 – 2013-02-12 16:58:45

+0

谢谢你的帮助!诀窍了。这个社区真棒:) – 2013-02-13 09:17:33