Php 表单显示代码中的一个错误

这是我的密码

<?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.";
}

?>

---

从这里删除最后一个点

 Comments:-     " . $comments .  

应该如此

 Comments:-     " . $comments;

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

现在您正在将另一行连接到$message；）

只需从行尾删除点（.），这样它就不会和下面的标题行连接。
应该是

 Comments:-     " . $comments;

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

您正在连接另一行
$headers=“From:”$电子邮件。“\r\n”
到
$message
字符串
您的
$message
声明应以
结尾
，而不是
，它在末尾包含
$headers
变量。