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我正在尝试使用以下代码通过我的网站处理客户联系。它不工作并且错误被捕获。我收到消息“抱歉，出现意外错误。请稍后再试”。

html 表单和 php 处理程序包括在下面。两者都在单独的页面中，分别是contact.php 和contact_handle.php。

形式：

  <div id="contact-form">
                <form method="post" action="contact_handle.php">

                    <div class="field">
                        <label>Name:</label>
                        <input type="text" name="name" class="text" />
                    </div>

                    <div class="field">
                        <label>Email: <span>*</span></label>
                        <input type="text" name="email" class="text" />
                    </div>

                    <div class="field">
                        <label>Subject: <span>*</span></label>
                        <input type="text" name="subject" class="text" />
                    </div>

                    <div class="field">
                        <label>Mobile No: <span>*</span></label>
                        <input type="text" name="mobile" class="text" />
                    </div>

                    <div class="field">
                        <label>Message: <span>*</span></label>
                        <textarea name="message" class="text textarea" ></textarea>
                    </div>

                    <div class="field">
                        <input type="button" id="send" value="Send Message"/>
                        <div class="loading"></div>
                    </div>

                </form>
</div

现在处理程序是：

<?php

//Retrieve form data. 
//GET - user submitted data using AJAX
//POST - in case user does not support javascript, we'll use POST instead
$name = ($_GET['name']) ? $_GET['name'] : $_POST['name'];
$email = ($_GET['email']) ?$_GET['email'] : $_POST['email'];
$subject = ($_GET['subject']) ?$_GET['subject'] : $_POST['subject'];
$mobile = ($_GET['mobile']) ?$_GET['mobile'] : $_POST['mobile'];
$message = ($_GET['message']) ?$_GET['message'] : $_POST['message'];

//flag to indicate which method it uses. If POST set it to 1
if ($_POST) $post=1;

//Simple server side validation for POST data, of course, you should validate the email
if (!$name) $errors[count($errors)] = 'Please enter your name.';
if (!$email) $errors[count($errors)] = 'Please enter your email.'; 
if (!$subject) $errors[count($errors)] = 'Please enter your subject.';
if (!$mobile) $errors[count($errors)] = 'Please enter your mobile No.';
if (!$message) $errors[count($errors)] = 'Please enter your message.'; 

//If the errors array is empty, send the mail
if (!$errors) {

    // ====== Your mail here  ====== //
    $to = 'fsa@gmail.com';

    // Sender
    $from = $name . ' <' . $email . '>';

    //subject and the html message
    $subject = 'Message from www.iso.com contact form'; 
    $message = '
    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" 
    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
    <html xmlns="http://www.w3.org/1999/xhtml">
    <head></head>
    <body>
    <table>
        <tr><td>Name:</td><td>' . $name . '</td></tr>
        <tr><td>Email:</td><td>' . $email . '</td></tr>
        <tr><td>Subject:</td><td>' . $subject . '</td></tr>
        <tr><td>Mobile No.:</td><td>' . $mobile . '</td></tr>
        <tr><td>Message:</td><td>' . nl2br($message) . '</td></tr>
    </table>
    </body>
    </html>';

    // Send the mail
    $result = sendmail($to, $subject, $message, $from);


    //if POST was used, display the message straight away
    if ($_POST) {
        if ($result) echo 'Thank you! We have received your message.';
        else echo 'Sorry, unexpected error. Please try again later';

    //else if GET was used, return the boolean value so that 
    //ajax script can react accordingly
    //1 means success, 0 means failed
    } else {
        echo $result;   
    }

// If the errors array has values
} else {}


// Simple mail function with HTML header
function sendmail($to, $subject, $message, $from) {
    $headers = "MIME-Version: 1.0" . "\r\n";
    $headers .= "Content-type:text/html;charset=iso-8859-1" . "\r\n";
    $headers .= 'From: ' . $from . "\r\n";

    $result = mail($to,$subject,$message,$headers);

    if ($result) return 1;
    else return 0;
}

?>

我希望你能帮助发现我错过的错误。

非常感谢 ：）

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2 回答 2

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该mail()功能仅与安装在服务器上的邮件传输一样可靠。在这种情况下，您应该检查服务器的邮件日志以获取有关邮件未被接受的原因的信息。

此外，如果您不确定信息是否会通过，$_GET或者$_POST您可以简单地使用$_REQUEST两者的混搭，再加上$_COOKIE记忆对我有用。

于 2013-02-11T17:04:59.307 回答
0

检查服务器的 PHP 错误日志（具体位置取决于您的服务器设置）。它将为您提供更多信息。

也就是说，根据您在这里的信息，我敢猜测您的呼叫sendmail()失败了。最可能的原因是您的服务器上未安装 PHP 使用的邮件发送器。

您需要确保设置了邮件发送程序，或者将 SMTP 与另一个邮件客户端（例如 gmail 帐户）一起使用来进行发送。

于 2013-02-11T17:06:01.593 回答