带有变量的php邮件答案

【问题标题】：php mail with variables带有变量的php邮件
【发布时间】：2011-10-09 17:38:47
【问题描述】：

我正在尝试设置一个邮件脚本，它将首先从 mysql 运行一个简单的选择，并在消息中使用这些数组变量。但是所有的变量都没有输出到消息体，只有一行变量。这是我的脚本：

    $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);

感谢任何人愿意分享的任何见解。谢谢， --马特

【问题讨论】：

标签： php mysql arrays email

【解决方案1】：

那是因为你一直在覆盖变量，所以它只会得到最后一个。您可能只想在 while 循环中构建 $message 变量。这将发送 1 封包含所有内容的电子邮件

 $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);

【讨论】：

你在某处有一些引号错误，并且消息会不断被重写 - 所以这也是错误的......

【解决方案2】：

那是因为您在循环中分配变量，但在循环外部的 $message 变量中使用它们。因此，您的 $message 仅包含最后一行/记录中的项目。尝试在 while 循环内的 $message 变量中移动和附加值。

应该是这样的

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()

希望对您有所帮助。

【讨论】：

【解决方案3】：

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" }

并删除循环下的另一个 $message 变量。注意 .在 $message 中的 = 之前。这告诉 php 不断添加到 $message

【讨论】：

【解决方案4】：

试试这个：

 $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。

【讨论】：

那将是一位不开心的管理员 :-)
@jeroen -- 那不是我的问题^_^（或者是......？）

【解决方案5】：

您正在覆盖循环中的变量，因此$name 等将包含循环后最后一行中找到的值。

你可以做的是在循环中构造你的消息：

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

【讨论】：