Question

我正在尝试设置一个邮件脚本，它将首先从mysql中运行一个简单的选择，并在消息中使用这些数组变量。但是，所有变量都不会输出到消息体，只有一行变量。这是我的剧本：

    $sql1 = "SELECT * FROM videos WHERE checked_out = '1'"; 
    $result1 = $dbLink->query($sql1); 
              while($row1 = $result1->fetch_assoc()) {
    $name = $row1['name'];
    $tape_no = $row1['tape_no'];
    $member_name = $row1['member_name'];
    $date_out = date("F j, Y", strtotime($row1['date_out']));
              }

//email function to administrator
$to = "nouser@mail.com";
$subject = "Daily Video Rental Summary";
$message = "$name $tape_no $date_out $member_name
======================================================================
PLEASE DO NOT REPLY TO THIS MESSAGE, AS THIS MAILBOX IS NOT MONITORED
======================================================================";
$from = "no_replies_please@mail.com";
$headers = "From:" . $from;
mail($to,$subject,$message,$headers);

感谢任何有关此人分享的见解。谢谢， --Matt

Answer 1

请改为尝试：

 $sql1 = "SELECT * FROM videos WHERE checked_out = '1'";
 $result1 = $dbLink - > query($sql1);
 while ($row1 = $result1 - > fetch_assoc()) {
     $name = $row1['name'];
     $tape_no = $row1['tape_no'];
     $member_name = $row1['member_name'];
     $date_out = date("F j, Y", strtotime($row1['date_out']));
     //email function to administrator
     $to = "nouser@mail.com";
     $subject = "Daily Video Rental Summary";
     $message = "$name $tape_no $date_out $member_name
======================================================================
PLEASE DO NOT REPLY TO THIS MESSAGE, AS THIS MAILBOX IS NOT MONITORED
======================================================================";
     $from = "no_replies_please@mail.com";
     $headers = "From:".$from;
     mail($to, $subject, $message, $headers);
 }

这将每行发送一封邮件。

Answer 2

那是因为你一直在覆盖变量，所以它只得到最后一个。您可能希望通过执行在while循环中构建$message变量。这将发送1封电子邮件，包含所有内容

 $sql1 = "SELECT * FROM videos WHERE checked_out = '1'"; 
    $result1 = $dbLink->query($sql1); 
              while($row1 = $result1->fetch_assoc()) {
    $name = $row1['name'];
    $tape_no = $row1['tape_no'];
    $member_name = $row1['member_name'];
    $date_out = date("F j, Y", strtotime($row1['date_out']));
    $message .= "$name $tape_no $date_out $member_name"
              }

//email function to administrator
$to = "nouser@mail.com";
$subject = "Daily Video Rental Summary";
======================================================================
PLEASE DO NOT REPLY TO THIS MESSAGE, AS THIS MAILBOX IS NOT MONITORED
======================================================================";
$from = "no_replies_please@mail.com";
$headers = "From:" . $from;
mail($to,$subject,$message,$headers);

Answer 3

您在循环中覆盖变量，因此$name等将包含循环后最后一行中找到的值。

您可以做的是在循环中构建您的消息：

$message = '';
while($row1 = $result1->fetch_assoc()) {
    $message .= $row1[...];    // whatever you need
}

Answer 4

那是因为你在循环中分配变量，但是在$ message变量 outside 循环中使用它们。所以你的$ message只包含最后一行/记录中的项目。尝试在while循环内的$ message变量中移动和追加值。

所以它可能是

while($row1 = $result1->fetch_assoc()) {
    //assign vars here
    $message .= "$name $tape_no $date_out $member_name\n";
}
$message = "$message
======================================================================
PLEASE DO NOT REPLY TO THIS MESSAGE, AS THIS MAILBOX IS NOT MONITORED
======================================================================";

//assign other variables
//mail()

希望有所帮助。

Answer 5

试试这个：

 $message = NULL; 
$sql1 = "SELECT * FROM videos WHERE checked_out = '1'"; 
$result1 = $dbLink->query($sql1); 
while($row1 = $result1->fetch_assoc()) { 
$name = $row1['name']; 
$tape_no = $row1['tape_no']; 
$member_name = $row1['member_name']; 
$date_out = date("F j, Y", strtotime($row1['date_out']));
 $message .= "$name $tape_no $date_out $member_name" }

并删除循环下的其他$ message变量。请注意。 $消息中的=之前。这告诉php不断添加$ message

带有变量的php邮件

5 个答案:

这将每行发送一封邮件。