Question

我的联系表格不愿意发送电子邮件。没有给出错误，但也没有收到邮件（我不在本地服务器上工作）。

我使用面向对象的方式，而不是最简单的方式，仍然可以工作..我的猜测是mail（）函数根本没有被触发，但我不知道理解为什么。

非常感谢所有帮助！

contact.php

<?php

include_once('classes/sendmail.php');

if(isset($_POST['sendmessage'])){
    try {
        $e = new Email();
        $e->Name = $_POST['name'];
        $e->Email = $_POST['email'];
        $e->Message = $_POST['message'];
        $e->Send();
    }

    catch(Exception $e){
        $error = $e->getMessage();
    }
}

if(isset($error)){
    echo $error;
}
?>


<!doctype html>
<body>


            <form name="contactform" action="" method="post" enctype="text/plain">
                    <label for="name">Name</label>
                    <input type="text" name="name" id="name" class="name" />

                    <label for="email">E-mail</label>
                    <input type="text" name="email" id="email" class="email" />

                    <label for="message">Message</label>
                    <textarea id="message" name="message" ></textarea>

                <input type="submit" class="send" value="Verzend" name="sendmessage" />
            </form>

类/ sendmail.php

<?php

class Email{

    private $m_sName;
    private $m_sEmail;
    private $m_sMessage;

    public function __set($p_sProperty, $p_vValue){
        switch($p_sProperty){

            case 'Name':
                if($p_vValue != '') {
                    $this->m_sName = $p_vValue;
                }
                else{
                    throw new Exception("No name found");
                }
                break;

            case 'Email':
            if($p_vValue != '') {
                    $this->m_sEmail = $p_vValue;
                }
                else{
                    throw new Exception("No email adress found");
                }
                break;

            case 'Message':
                if($p_vValue != '') {
                    $this->m_sMessage = $p_vValue;
                }
                else{
                    throw new Exception("No message found");
                }
                break;
        }
    }

    public function __get($p_sProperty){
        switch($p_sProperty){
            case 'Name':
                return $this->m_sName;
                break;

            case 'Email':
                return $this->m_sEmail;
                break;

            case 'Message':
                return $this->m_sMessage;
                break;
        }
    }

    public function Send(){

        $email_to = "<my Email>";
        $email_subject = "<my subject>";

        $email_message = "Name: ".$this->m_sName."\n";
        $email_message .= "Email: ".$this->m_sEmail."\n";
        $email_message .= "Message: ".$this->m_sMessage."\n";

        $headers = 'From: '.$this->m_sEmail."\r\n".
        'Reply-To: '.$this->m_sEmail."\r\n" .
        'X-Mailer: PHP/' . phpversion();

        mail($email_to, $email_subject, $email_message, $headers);

    }

}

?>

编辑：已经有很多关于联系表格的问题，但是没有人使用OOP似乎和我一样有问题。

Answer 1

您的问题不在PHP类中。它不会抛出错误，因为它没有收到任何$_POST数据。您需要从enctype="text/plan"中移除<form>。

完成后应该是这样的：

<form name="contactform" action="" method="post">

Answer 2

在测试表单之前，您是否已经设置了smtp / sendmail？

您可以使用Library PHPMailer

示例：

function send_email($subject, $from_mail, $message) {

            $mail = new PHPMailer();
            $mail->IsSMTP();  // send via SMTP
            $mail->Host     = setting_db('SMTPHOST');   // SMTP servers
            $mail->SMTPAuth = true;                  // turn on 
            $mail->Username =  setting_db('SMTPUSER');  // SMTP username
            $mail->Password =  setting_db('SMTPPASSWORD'); // SMTP password
            $mail->From     =  setting_db('SMTPUSER');  // // SMTP username from
            $mail->FromName =  "Your Name";
            $mail->AddAddress($from_mail); // 
            $mail->Subject  = $subject;
            $mail->Body     = $message;
            if ($mail->send()) {
                return "OK!";
            } else {
                return "No!". $mail->ErrorInfo;
            }

}

并更新文件contact.php

send_email($_POST["subject"], $_POST["email"], $_POST["message"]);

替换

$e = new Email();
$e->Name = $_POST['name'];
$e->Email = $_POST['email'];
$e->Message = $_POST['message'];
$e->Send();

PHP面向对象的联系表单

2 个答案: