如何使用变量发送消息

-1

$mail = new PHPMailer(true); 
$mail->IsSMTP(); 
$mail->SMTPDebug = 0; 
$mail->SMTPAuth = true; 
$mail->SMTPSecure = 'ssl'; 
// $mail->Host = "smtp.gmail.com"; 
$mail->Host = "mail.infysky.com"; 
$port = $mail->Port = 465; 
//echo "<br />"; 
$mail->IsHTML(false); 

$name  = "'" . $params["txtName"] . "'"; 
$email  = "'" . $params["txtEmailID"] . "'"; 
$department = "'" . $params["selDepartment"] . "'"; 
$designation = "'" . $params["selDesignation"] . "'"; 
$leave  = "'" . $params["selLeave"] . "'"; 
$dateFrom = "'" . $params["txtDate1"] . "'"; 
$dateTo  = "'" . $params["txtDate2"] . "'"; 
$reason  = "'" . $params["txtReason"] . "'"; 

$mail->SetFrom("[email protected]", 'infy SKY'); 
$mail->Subject = "90%"; 
$mail->Body = "Message for leave apply \n NAME:  $name;\n E-MAIL  
$email;\n DEPARTMENT: $department; \n DESIGNATION: $designation;\n LEAVE 
TYPE: $leave;\n DATE FROM: $dateFrom;\n DATE TO:  $dateTo;\n REASON:  
$reason; "; 

$mail->AddAddress(); //To Address 
$mail->Send();

如何从PHP Mailer中的AddAddress变量发送消息。在AddAddress中，我想发送动态消息到输入的客户端。如何使用变量发送消息

来源

2017-10-04 Karthik

邮件顺利通过这段代码发送只是我想$电子邮件发送的AddAddress – Karthik

BRO I GOT解决方案 – Karthik

你试过$mail->addAddress($email)？见https://github.com/PHPMailer/PHPMailer

来源

2017-10-04 07:41:32

是的，我试过，但我得到电子邮件地址无效 – Karthik

你可以打印出来或的var_dump（）的内容你'$ email'变量？ –

我正在通过var_dump（）传递的电子邮件地址，但仍然收到此错误消息致命错误：带有消息'Unvalidaught address'的未捕获异常'phpmailerException'：[email protected]'' – Karthik

如何使用变量发送消息

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