如果$ from中没有使用gmail,我的表单不会发送到邮件

时间:2019-05-06 09:36:57

标签: php forms

我已经设置了一个PHP表单,但是不管我尝试什么,$ from必须是gmail,但奇怪的是$ from必须是一个gmail帐户。需要它使用任何帐户作为info @ site服务器上的一个帐户。在$ from中使用gmail可以完美工作,这没有任何意义,因为我相信它甚至不必作为真实地址存在,sento变量是唯一需要实际地址的变量。对多封邮件进行了测试,不仅对注册到该站点的邮件进行了测试,甚至对hotmail和Outlook和mail.com进行了测试,但只有gmail可以工作。

<?php
/*
*  CONFIGURE EVERYTHING HERE
*/

// an email address that will be in the From field of the email.
$from = 'info@site.com';

// an email address that will receive the email with the output of the form
$sendTo = 'site@email.com';

// subject of the email
$subject = 'New message From Web Form';

// form field names and their translations.
// array variable name => Text to appear in the email
$fields = array('name' => 'Name', 'surname' => 'Surname', 'phone' => 'Phone', 'email' => 'Email', 'message' => 'Message','post' => 'Postal Code'); 

// message that will be displayed when everything is OK :)
$okMessage = 'Contact form successfully submitted. Thank you, we will get back to you soon!';

// If something goes wrong, we will display this message.
$errorMessage = 'There was an error while submitting the form. Please try again later';

/*
*  LET'S DO THE SENDING
*/

// if you are not debugging and don't need error reporting, turn this off by error_reporting(0);
error_reporting(E_ALL & ~E_NOTICE);

try
{

if(count($_POST) == 0) throw new \Exception('Form is empty');

$emailText = "You have a new message from your contact 
form\n=============================\n";

foreach ($_POST as $key => $value) {
    // If the field exists in the $fields array, include it in the email 
    if (isset($fields[$key])) {
        $emailText .= "$fields[$key]: $value\n";
    }
}

// All the neccessary headers for the email.
$headers = array('Content-Type: text/plain; charset="UTF-8";',
    'From: ' . $from,
    'Reply-To: ' . $from,
    'Return-Path: ' . $from,
);

// Send email
mail($sendTo, $subject, $emailText, implode("\n", $headers));

$responseArray = array('type' => 'success', 'message' => $okMessage);
}
catch (\Exception $e)
{
$responseArray = array('type' => 'danger', 'message' => $errorMessage);
}


// if requested by AJAX request return JSON response
if (!empty($_SERVER['HTTP_X_REQUESTED_WITH']) && 
strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest') {
$encoded = json_encode($responseArray);

header('Content-Type: application/json');

echo $encoded;
}
// else just display the message
else {
echo $responseArray['message'];
}

0 个答案:

没有答案