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最近,我使用 php 脚本从在线联系表单发送电子邮件时发生了一些奇怪的事情,我只是想知道是否有人可以对这个问题有所了解。

我有一个在多个网站上使用的 php 脚本,它一直运行良好,但出于某种奇怪的原因,我尝试在一个网站上使用它,但它无法正常工作。

我试着摆弄它,最终意识到它与以下代码部分有关:

这是通常可以正常工作的原始代码部分,但由于某种原因无法正常工作:

$to = 'My Name <info@mydomain.com>';

然后我删除了名称位,因此代码如下所示:

$to = 'info@mydomain.com';

现在它通过确定发送电子邮件。

正如我所说,顶级代码通常可以正常工作,所以有什么想法为什么这次我必须更改代码才能使其正常工作?

任何可能的解释都会很棒:o)

这是完整的代码:

<?php

require("is_email.php"); // email validation function

//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'];  
$telephone = ($_GET['telephone']) ?$_GET['telephone'] : $_POST['telephone'];
$address = ($_GET['address']) ?$_GET['address'] : $_POST['address'];
$enquiry = ($_GET['enquiry']) ?$_GET['enquiry'] : $_POST['enquiry'];
$calculation = ($_GET['calculation']) ?$_GET['calculation'] : $_POST['calculation']; 

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

//Server side validation for POST data
if (!$name) $errors[count($errors)] = 'Please click back and enter your name.';  
if (!$email) $errors[count($errors)] = 'Please click back and enter your email.'; 
else if (!is_email($email)) $errors[count($errors)] = 'Please click back as you may have entered an invalid email address.';
if (!$telephone) $errors[count($errors)] = 'Please click back and enter your telephone number.';
if (!$address) $errors[count($errors)] = 'Please click back and enter your address.';
if (!$enquiry) $errors[count($errors)] = 'Please click back and enter your enquiry.';
if ($calculation != '14') $errors[count($errors)] = 'Please click back and check you have correctly answered the simple calculation (in number format).';

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

    //recipient - change this to your name and email  
    $to = 'info@mydomain.com';     
    //sender  
    $from = $name . ' <' . $email . '>';

    //subject and the html message  
    $subject = 'Website Enquiry: ' . $name;   
    $email_body = '  
    <!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 cellpadding="5" style="color:#757575;">  
        <tr><td style="color:#3b5998;">Name: </td><td>' . $name . '</td></tr>  
        <tr><td style="color:#3b5998;">Email: </td><td>' . $email . '</td></tr>  
        <tr><td style="color:#3b5998;">Telephone: </td><td>' . $telephone . '</td></tr>
        <tr valign="top"><td style="color:#3b5998;">Address: </td><td>' . nl2br($address) . '</td></tr>
        <tr valign="top"><td style="color:#3b5998;">Enquiry: </td><td>' . nl2br($enquiry) . '</td></tr>  
    </table>  
    </body>  
    </html>';

    //send the mail  
    $result = sendmail($to, $subject, $email_body, $from);

    //if POST was used, display the message straight away  
    if ($_POST) {  
        if ($result) echo 'Thank you! We have received your message.<br /><br /><a href="../enquiry_form.html">OK</a>';  
        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 {

    //display the errors message  
    for ($i=0; $i<count($errors); $i++) echo $errors[$i] . '<br />';    
    exit;  
}

//Simple mail function with HTML header  
function sendmail($to, $subject, $email_body, $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,$subject,$email_body,$headers);  

    if ($result) return 1;  
    else return 0;  
}  
?>  
4

3 回答 3

0

主机可以根据他们的要求sendmail以及sendmail他们使用的软件而有所不同。您可以与您的托管公司讨论这个问题,看看是否有任何警告可以让它发挥作用,或者他们是否知道更好的格式。

于 2010-09-08T14:32:14.897 回答
0

用这个:

$mailFrom = '"My Name" <info@mydomain.com>';

FROM 标签是双引号和一个空格和<email>

于 2010-09-08T14:42:17.307 回答
0

我编辑了我最初的问题以解释我是如何克服这个问题的。

于 2011-12-23T11:26:36.030 回答