php脚本正在输出php代码,不知道为什么

Question

我有一个简单的模态对话框,我在自己的linux服务器上开发,运行php 5.3.脚本(如下所示)在我的服务器上运行正常.但是,我将它移动到我的客户端的linux服务器,而不是回显它显然应该做的文本/ html,它从>(大于)字符回显所有实际的PHP代码.有谁知道为什么它会回应实际代码？有没有导致这个的php.ini设置？或两个设置中的文件编码差异？

<?php


$to_email = 'myname@myemail.com';
 $link = $_GET['link'];
 if(!$link){
  echo '<p>Have a suggestion?<br />Enter the URL below!</p>';
 }else if(strlen($link) > 256 || !preg_match('/^(http:\/\/)?subdomain\.somesite\.com\/(somedir\/)?anotherdir\/(.+)/',$link) && !preg_match('/^(http:\/\/)?somedomain2\.com\/somedir2\/(.+)/',$link)){
  echo '<p class="error">Whoops, the URL entered doesn\'t <br />match the criteria.</p>';
 }else{
  $link = str_replace("\n.", "\n..", $link);
  if(!preg_match('/^http:\/\//',$link)){
   $link = 'http://'.$link;
  }
  mail($to_email, 'New URL Suggestion', "A new URL has been suggested from your site:\n\n".$link,"From: ".$to_email."\r\n");
  echo '<p>Thank you for submitting this URL! <br />It should be live within 24 hours.</p>';
 }
?>

我客户端服务器上的结果是:

256 || !preg_match('/^(http:\/\/)?subdomain\.somesite\.com\/(somedir\/)?anotherdir\/(.+)/',$link) &&
!preg_match('/^(http:\/\/)?somedomain2\.com\/somedir2\/(.+)/',$link)){ echo '
Whoops, the URL entered doesn\'t 
match the criteria.

'; }else{ $link = str_replace("\n.", "\n..", $link);
if(!preg_match('/^http:\/\//',$link)){ $link = 'http://'.$link; } mail($to_email,
'New URL Suggestion', "A new URL has been suggested from your site:\n\n".$link,"From:
".$to_email."\r\n"); echo '
Thank you for submitting this URL! 
It should be live within 24 hours.

'; } ?>

Answer 1

听起来像其他服务器没有配置为运行PHP.在配置中是否有这样的行？

AddType application/x-httpd-php .php .html .htm