注意：未定义索引：HTTP_REFERER

Question

注意：未定义的索引：HTTP_REFERER

$http_referer = $_SERVER ['HTTP_REFERER']

我从教程中使用了这个，看起来没问题

而且代码是从包含文件中调用它的

我应该改变什么？

我添加了print_r（$ _SERVER）;现在页面给了我这个

 Array ([UNIQUE_ID] => UoSxWa56310AAAwUckIAAAAA
        [HTTP_HOST] => movafaghha.com
        [HTTP_COOKIE] => __utma=210711305.58608218.1372977010.1372977010.1372977010.1; __utmz=210711305.1372977010.1.1.utmcsr=who.is|utmccn=(referral)|utmcmd=referral|utmcct=/whois/movafaghha.com; PHPSESSID=83eb0e2ae7ebe4b6c2eeb071d9f5de71
        [HTTP_X_REAL_IP] => 109.109.41.81
        [HTTP_X_FORWARDED_HOST] => movafaghha.com
        [HTTP_X_FORWARDED_SERVER] => movafaghha.com
        [HTTP_X_FORWARDED_FOR] => 109.109.41.81
        [HTTP_CONNECTION] => close 
        [HTTP_CACHE_CONTROL] => max-age=0 
        [HTTP_ACCEPT] => text/html,application/xhtml+xml,application/xml;q=0.9,image/webp,*/*;q=0.8 
        [HTTP_USER_AGENT] => Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/31.0.1650.48 Safari/537.36 
        [HTTP_ACCEPT_LANGUAGE] => en-US,en;q=0.8,fa;q=0.6 
        [PATH] => /sbin:/usr/sbin:/bin:/usr/bin
        [SERVER_SIGNATURE] =>
        [SERVER_SOFTWARE] => Apache 
        [SERVER_NAME] => movafaghha.com 
        [SERVER_ADDR] => 174.122.223.93 
        [SERVER_PORT] => 80 
        [REMOTE_ADDR] => 109.109.41.81 
        [DOCUMENT_ROOT] => /home/memarest/public_html/movafaghha.com 
        [SERVER_ADMIN] => [e-mail] 
        [SCRIPT_FILENAME] => /home/memarest/public_html/movafaghha.com/tutorials/login200/register.php 
        [REMOTE_PORT] => 49737 
        [GATEWAY_INTERFACE] => CGI/1.1 
        [SERVER_PROTOCOL] => HTTP/1.0 
        [REQUEST_METHOD] => GET 
        [QUERY_STRING] => 
        [REQUEST_URI] => /tutorials/login200/register.php 
        [SCRIPT_NAME] => /tutorials/login200/register.php 
        [PHP_SELF] => /tutorials/login200/register.php 
        [REQUEST_TIME_FLOAT] => 1384427865.54 
        [REQUEST_TIME] => 1384427865 
        [argv] => Array ( ) 
        [argc] => 0 
    ) 

我修改了代码，但仍无法回显所有的字段都是必填的。

<?php

ini_set("display_errors", true);
error_reporting(E_ALL);

require 'core.inc.php';






if(!loggedIn()) {

//check mikunim ke tamame field ha dar form vojod darand va set shudan

if(isset($_POST['username'])&&isset($_POST['password'])&&isset($_POST['password_again'])&&isset($_POST['firstname'])&&isset($_POST['surename'])) {

    $username = $_POST['username'];
    $password = $_POST['password'];
    $password_again = $_POST['password_again'];
    $firtsname = $_POST['firstname'];
    $surename = $_POST['surename'];



    //HALA CHECK MIKUNIM KHALI HASTAND YA NA

    if(!empty($username)&&!empty($password)&&!empty($password_again)&&!empty($firstname)&&!empty($surename)){
        echo 'ok' ;

    } else {

        echo ' All fields are required';

    }

}




?>

<form action="register.php" method="POST">
Username:<br> <input type="text" name="username"><br> <br>
Password:<br> <input type="password" name="password"><br><br>
Password again:<br> <input type="password" name="password_again"><br><br>
Firstname:<br> <input type="text" name="firstname"><br><br>
Surname:<br> <input type="text" name="surename"><br><br>
<input type="submit" value="register">



</form>

<?php

} elseif (loggedIn()) {

    echo 'you \'re already logged in';
    }

?>

现在添加了

"e"

页面显示“所有字段都是必填项”

但即使填写了所有字段，消息也不会改变。

- Mojtaba

---

你应该测试这个变量是否存在。如果你不是从另一个页面跳转而是直接在地址栏中输入链接进入页面，那么就不会有引用来源... - Laurent S.

Answer 1

HTTP_REFERER不能保证由客户端发送：

该页面（如果有的话）引用了用户代理到当前页面的地址。这是由用户代理设置的。并非所有用户代理都会设置此值，有些提供修改HTTP_REFERER的功能。简而言之，它并不是完全可信的。

在您的情况下，很明显它没有被发送，因此您能做的就只有

if(isset($_SERVER['HTTP_REFERER'])) {
  //do what you need to do here if it's set    
   }
else
{
   //it was not sent, perform your default actions here
}

- Stephen Byrne

---

嗨，我认为原始问题可能要求阻止一些垃圾网站（至少在我的情况下）。在上面的解决方案中，将在所有情况下执行操作。这是我所做的：如果（isset（$ _SERVER ['HTTP_REFERER']）&& $ _SERVER ['HTTP_REFERER'] == 'www.spamwebsite.com'）{ // action}。 - coder

Answer 2

if (isset($_SERVER['HTTP_REFERER'])) {$THE_REFER=$_SERVER['HTTP_REFERER']}

- Mahdi Jazini

---

isset 方法返回一个布尔值。你的 $THE_REFER 变量将是 "true" 或 "false"，而不是 referer 头的实际值。 - Jeff Kilbride

@JeffKilbride 谢谢您的评论。这篇文章是两年前发布的...我不知道为什么会发这样一个不合逻辑的帖子！我很震惊！我已经编辑过了。再次感谢您。 - Mahdi Jazini

Answer 3

未定义索引意味着数组键未设置，请执行以下操作：

var_dump($_POST); die(); 

在引发错误的那行代码之前，查看一下你是否试图获取一个不存在的数组键。

- ashish

Answer 4

正确的引用方式是：

$my_referer = isset($_POST['referer']) ? trim($_POST['referer']) : (isset($_SERVER['HTTP_REFERER']) ? base64_encode($_SERVER['HTTP_REFERER']) : false);

- rn consulting