Question

这是我的代码

<?php
$email=$_POST['email'];
$name=$_POST['name'];
$comments=$_POST['comments'];

$to = "my@address.com";
$subject = "Comments";
$message = 
"
 Name:-             " . $name .     "\r\n   
 Email:-            " . $email .    "\r\n 
 Comments:-     " . $comments .                  

$headers = "From:" . $email . "\r\n";

if(@mail($to,$subject,$message,$headers))
{
  print "<script>document.location.href='http://thesite.org/docs/tkx.html';</script>";

}
else
{
  echo "Error! Please try again.";
}

?>

但由于我无法检测到的原因，当数据传送到我的电子邮件时，其中包含“评论”数据末尾的“电子邮件地址”数据

我做错了什么？

提前致谢。 OzWaz

Answer 1

从此处删除最后一个点

 Comments:-     " . $comments .

所以它应该是

 Comments:-     " . $comments;

现在你将另一行连接到$ message;）

Answer 2

$message = 
"
 Name:-             " . $name .     "\r\n   
 Email:-            " . $email .    "\r\n 
 Comments:-     " . $comments .

只需从行尾删除点（。），使其不与下面的标题行连接。它应该是

$message = 
"
 Name:-             " . $name .     "\r\n   
 Email:-            " . $email .    "\r\n 
 Comments:-     " . $comments;

Answer 3

$message = 
"
Name:-             " . $name .     "\r\n   
Email:-            " . $email .    "\r\n 
Comments:-     " . $comments;                  

$headers = "From:" . $email . "\r\n";

您正在将另一行$headers = "From:" . $email . "\r\n";引入$message字符串

表单中的错误显示代码

3 个答案: