2015-04-23 55 views
0

因此,我已经构建了此电子邮件表单,如果未提交名称或/和电子邮件,应该显示错误消息'名称和电子邮件是必需的'。但是,如果他们是我收到相同的消息。这怎么解决?PHP - 在没有错误时提交错误信息

<?php 
if(!isset($_POST['submit-enquiry'])) 
{ 
    //This page should not be accessed directly. Need to submit the form. 
    echo "error; you need to submit the form!"; 
} 
$name = $_POST['name']; 
$guest_email = $_POST['email']; 
$mobile = $_POST['mobile']; 
$message = $_POST['enquiry-message']; 

//Validate first 
if(empty($name)||empty($guest_email)) 
{ 
    echo '<script language="javascript">'; 
    echo 'alert("Name and email are mandatory")'; 
    echo '</script>'; 
    exit; 
} 

if(IsInjected($visitor_email)) 
{ 
    echo '<script language="javascript">'; 
    echo 'alert("Bad Email Value")'; 
    echo '</script>'; 
    exit; 
} 

$email_from = $guest_email;//<== update the email address 
$email_subject = "Enquiry from $name"; 
$email_body = "Name: $name. \n". "Mobile: $mobile .\n". "Message:  $message. \n"; 

$to = "my_email";//<== update the email address 
$headers = "From: $email_from \r\n"; 
$headers .= "Reply-To: $guest_email \r\n"; 
//Send the email! 
mail($to,$email_subject,$email_body,$headers); 
//done. redirect to thank-you page. 
header('Location: enquiry.php'); 

// 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; 
    } 
} 

?> 

的fprm的HTML:

<div id="enquiry-form"> 
        <form method="post" name="enquiry-form" action="" target="_self"> 
         <span class="short-input" id="name"> 
          <h6>name</h6> 
          <input type="text" name="name"> 
         </span> 
         <span class="short-input" id="mobile"> 
          <h6>mobile</h6> 
          <input type="text" name="mobile"> 
         </span> 
         <span class="long-input" id="email"> 
          <h6>e-mail</h6> 
          <input type="text" name="email"> 
         </span> 
         <span class="long-input" id="enquiry-message"> 
          <h6>enquiry</h6> 
          <textarea name="enquiry-message"></textarea> 
         </span> 

       </div> 
       <div id="contact-info"> 
        <h2>Contact Details</h2> 
       </div> 
       <button type="submit" id="submit-enquiry" name="submit-enquiry">send</button> 
       </form> 
+0

你能做一个$ _POST转储吗? 你确定你的表单没有使用GET而不是POST吗? 也许向我们展示表单的HTML? – Spode

回答

0

尝试

if($name=='' || $guest_email=='') 

在某些情况下空洞会导致意想不到的结果。我的经验让我说,对于表单验证,最好通过这种方式检查变量是否为空。然后你检查变量的内容(例如,看看它是否是一个有效的电子邮件),所以这对我来说足够安全。 如果你的代码仍然不起作用,那么有必要做进一步的调查。

+0

这不提供问题的答案。要批评或要求作者澄清,请在其帖子下方留言。 –

+0

你是对的。当我在电话中时,我错误地发布了我的答案,并且缺少部分文本。 :) –

0
if(empty(trim($name)) || empty(trim($guest_email)))