Question

我正在尝试创建一个函数，可以检查数据库中是否有活动状态为1的活动客户，他们应该接收电子邮件。对于电子邮件功能，我使用过PHP Mailer功能。

以下是我的剧本：

＆＃13;

<?php
include '../mailer/class.phpmailer.php';
$sqlx = mysql_query("SELECT * from `cust");

$numRows = mysql_num_rows($sqlx);
$mail_body = '';
while($row = mysql_fetch_array($sqlx))
{
// fetch email

$uid = $row["uid"];
$email = $row["email"];

$count = "20";
if($count <= '70')
{
$f_name = "abc";
$f_email = "abc@xyz.com";
$mail_body = "Hii message";
$subject = "Hi you got notificaiton";
$headers  = "From: abc <abc@xyz.com>";
$headers .= "Content-type: text/html\r\n";

function Send_mail($email, $subject, $headers, $mail_body, $f_email, $f_name)
{
	$mail = new PHPMailer(); 
	$Email = $email;
	$fname = $f_name;
	$femail = $f_email; 
	//==================smtp mail ===============================//
	$mail->IsSMTP(); // telling the class to use SMTP
	$mail->SMTPAuth = true;
	$mail->Port= 25; //  Sets the default SMTP server port.
	$mail->SMTPSecure= 'tls'; //  Options are "", "ssl" or "tls"
	$mail->Host = 'localhost'; // SMTP server	
	$mail->Username = 'abc@xyz.com';  // Sets SMTP username.	
	$mail->Password = '1234556';  //Sets SMTP password.	
	//=========================================================//	
		
	$fname = $f_name;
	$femail = $f_email; 						// email address of reciever	
	$mail->WordWrap = 50;                              // set word wrap	
	$mail->IsHTML(true);                               // send as HTML		
	$mail->Subject  =  $subject;						// subject of mail	
	$mail->Body     =  $mail_body;						// body of mail		
	$mail->Send();			return true;
}

$mail_result=Send_Mail($email, $subject, $headers, $mail_body,$f_email, $f_name);


}

}

?>

＆＃13;

但是我收到了错误：

您必须至少提供一个收件人电子邮件地址。致命错误：无法在第161行的C：\ xampp \ htdocs \ testing \ update.php中重新声明Send_mail（）（之前在C：\ xampp \ htdocs \ testing \ update.php：161中声明）

Answer 1

将你的函数放在循环之外。将你的代码结构改为此。

function Send_mail($email, $subject, $headers, $mail_body, $f_email, $f_name)
        {
            ................//code
            $mail->Send();          return true;
        }


     while($row = mysql_fetch_array($sqlx))//use mysqli or PDO
        {
       .......//code
        if($count <= '70')
         {
        .......//code
        $mail_result=Send_mail($email, $subject, $headers, $mail_body,$f_email, $f_name);
        }
        }

Answer 2

检查客户表，必须输入您要在其上发送电子邮件的电子邮件地址。并更改您的

$mail->Port= 587; 
$mail->SMTPSecure= 'tls'; 
$mail->Host = 'smtp.gmail.com'; 
$mail->Username = 'abc@gmail.com';//Valid Gmail address 
$mail->Password = '1234556';//Gmail password

如果仍然出现错误，则可以转到gmail帐户设置并允许安全的应用程序身份验证。

Answer 3

将您的功能置于循环之外。这是代码结构：

<?php
include '../mailer/class.phpmailer.php';
function Send_Mail($email, $subject, $headers, $mail_body, $f_email, $f_name)
{
    $mail = new PHPMailer(); 
    $Email = $email;
    $fname = $f_name;
    $femail = $f_email; 
    //==================smtp mail ===============================//
    $mail->IsSMTP(); // telling the class to use SMTP
    $mail->SMTPAuth = true;
    $mail->Port= 25; //  Sets the default SMTP server port.
    $mail->SMTPSecure= 'tls'; //  Options are "", "ssl" or "tls"
    $mail->Host = 'localhost'; // SMTP server   
    $mail->Username = 'abc@xyz.com';  // Sets SMTP username.    
    $mail->Password = '1234556';  //Sets SMTP password. 
    //=========================================================//   

    $fname = $f_name;
    $femail = $f_email;     // email address of reciever    
    $mail->WordWrap = 50;   // set word wrap    
    $mail->IsHTML(true);    // send as HTML     
    $mail->Subject  =  $subject;// subject of mail  
    $mail->Body     =  $mail_body;  // body of mail     
    $mail->Send();          
    return true;
}




$sqlx = mysql_query("SELECT * from `cust");

$numRows = mysql_num_rows($sqlx);
$mail_body = '';
while($row = mysql_fetch_array($sqlx))
{
// fetch email

$uid = $row["uid"];
$email = $row["email"];

$count = "20";
if($count <= '70')
{
$f_name = "abc";
$f_email = "abc@xyz.com";
$mail_body = "Hii message";
$subject = "Hi you got notificaiton";
$headers  = "From: abc <abc@xyz.com>";
$headers .= "Content-type: text/html\r\n";
$mail_result=Send_Mail($email, $subject, $headers, $mail_body,$f_email, $f_name);


}

}

＆GT;

无法重新声明Send_mail（）错误

3 个答案: