Question

我正在尝试使用php制作一个功能表单。最初我制作了一个“index.html”文件和另一个名为send.php的文件。但它没有用。我也尝试在send.php中插入“echo”，它有效！所以我将index.html转换为index.php，并在我关闭标签之前将php代码放在其中，但它没有用！所以我意识到错误是在代码中，但我从互联网复制这些行，我不知道PHP非常好。有人能帮我吗？这是代码：

字段的名称=“”是正确的，我查了一下。

<?php 
$name = ($_POST['name'];
$mail = ($_POST['mail'];
$budjet = ($_POST['budjet'];
$society = ($_POST['society'];
$message = ($_POST['message'];
$error = "Message sending failed";
$errorMessage = "Sorry your message can not be sent.";

//Validate first
if(empty($name)||empty($mail)||empty($budjet)||empty($society)||empty($message)) 
{
    echo "Please fill al the fields";
    header('Location: index.html');
}
//validate against any email injection attempts
if(IsInjected($email))
{
    echo "Bad email value";
    header('Location: index.html');
}

$msg = "Name : $name \r\n"; 
$msg = "Mail : $mail \r\n";
$msg = "Budjet: $budjet \r\n";
$msg = "Society: $society \r\n";
$msg = "Message : ".stripslashes($_POST['message'])."\r\n\n";
$msg = "User information \r\n"; 
$msg = "User IP : ".$_SERVER["REMOTE_ADDR"]."\r\n"; 
$msg = "Browser info : ".$_SERVER["HTTP_USER_AGENT"]."\r\n"; 
$msg = "User come from : ".$_SERVER["SERVER_NAME"];

$recipient = "myemail@mail.com"; // Change the recipient email adress to your adrees  
$sujet =  "Mail da Contact di Hindeto";
$mailheaders = "From: $email\r\nReply-To: $email\r\nReturn-Path: $email\r\n";

if (!$error){

        $sending = mail($recipient, $sujet, $msg, $mailheaders); 

        if ($sending) {
                // If the message is sent we output a string to use it 
                echo "Sent"; 
            } else {
                // Display Error Message
                echo $errorMessage; 
            }
    } else {
        echo $error; // Display Error Message
    }


// Function to validate against any email injection attempts
function IsInjected($str)
{
  $injections = array('(\n+)',
              '(\r+)',
              '(\t+)',
              '(%0A+)',
              '(%0D+)',
              '(%08+)',
              '(%09+)'
              );
  $inject = join('|', $injections);
  $inject = "/$inject/i";
  if(preg_match($inject,$str))
    {
    return true;
  }
  else
    {
    return false;
  }
}
?>

Answer 1

您在(

之前的前5行删除了$_POST时出现语法错误

$name = $_POST['name'];
$mail = $_POST['mail'];
$budjet = $_POST['budjet'];
$society = $_POST['society'];
$message = $_POST['message'];

之后 header('Location: index.html');和header('Location: index.html');打印exit();

header('Location: index.html');
 exit();

Answer 2

除Jetawe外，您还必须在验证声明中使用isset（）。

$name = ((!isset($_POST['name']) || empty($_POST['name'])) ? null:$_POST['name']);

无法获取php输出表单

2 个答案: